Trigonometry 

AND 

Stereographic  Projections 


Trigonometry 


AND 


Stereographic  Projections 


(REVISED) 


PREPARED  FOR  THE  USE  OF  THE 

MIDSHIPMEN  AT  THE  UNITED  STATES 

NAVAL  ACADEMY 


BY 

STIMSON  J.  BROWN 

Professor  of  Mathematics,  United  States  Navy 


1913 


Copyright,  1913,  by 

RALPH  EARLE 

Secy,  and  Treas.  U.  S.  Naval  Institute 


(gafttmore  (p 

BALTIMORE,  MD.,  U.  S.  A. 


PKEFACE. 

I  have  attempted  in  the  work  here  presented  to  give  a  thor- 
oughly practical  treatment  of  the  subject  of  trigonometry,  adapted 
to  the  requirements  of  the  course  of  study  of  the  Naval  Academy. 

The  chapter  on  stereographic  projections  represents  largely 
the  methods  of  Professor  W.  W.  Hendrickson,  U.  S.  Navy,  by 
whose  courtesy  some  of  the  articles  and  plates  have  been  taken 
directly  from  his  work  on  that  subject. 

S.  J.  B. 

JANUABY  18,  1913. 


3G0378 


CONTENTS. 

PLANE  TRIGONOMETRY. 

CHAPTER  I.  PAGE 

Angles  and   their  Measure 1 

Sexagesimal  Units   2 

Circular  Units   3 

Compass   Units    5 

CHAPTER  II. 

Definitions  of  Trigonometric  Functions . 8 

Complementary   Functions    11 

Fundamental  Formulae   12 

Line  Representation  of  Functions 15 

Reduction  of  Functions  to  the  First  Quadrant 16 

Periodicity  of  Trigonometric  Functions 20 

Inverse   Functions    20 

CHAPTER  III. 

Explanation  of  the  Use  of  Trigonometric  Tables 23 

Graphs  of  Trigonometric  Functions 31 

CHAPTER  IV. 

Solution  of  Plane  Right  Triangles 35 

Polar   Coordinates    38 

Graphs,   Polar   Coordinates 40 

CHAPTER  V. 

Functions  of  the  Sum  or  Difference  of  Two  Angles,  Double-angle, 

and  Half-angle  43 

Sum  of  Two  Inverse  Functions..  48 


viii  CONTENTS. 

CHAPTER  VI.  PAGE 

Formulae  for  Oblique  Triangles 50 

The  Sine  Formula 50 

The  Cosine  Formula 51 

The  Tangent  Formula 51 

Angles  in  Terms  of  the  Sides 52 

Area  of  a  Triangle 53 

CHAPTER  VII. 

Solution  of  Oblique  Triangles 55 

Radius  of  Circumscribed  and  Inscribed  Circles 61 

CHAPTER  VIII. 

Graphic  Representation  of  Complex  Numbers 63 

De  Moivre's  Theorem 65 

Functions  of  Multiple  Angles 66 

Series  for  Sine  and  Cosine 67 

Roots   of   Imaginaries 70 


SPHERICAL  TRIGONOMETRY. 

CHAPTER  IX. 

General  Formulae   71 

The  Three  Sides  and  an  Angle 72 

The  Three  Angles  and  a  Side 73 

The  Sine  Formula 74 

The  Half-angles  in  Terms  of  the  Sides 76 

The  Half-sides  in  Terms  of  the  Angles 77 

Gauss's   Equations    78 

Napier's  Analogies 79 

CHAPTER  X. 

Solution  of  Spherical  Right  Triangles 80 

Napier's  Rules  81 

The  Law  of  Quadrants..  82 


CONTENTS.  ix 

CHAPTER  X. — Continued.  PAGE 

The  Double  Solution 83 

Triangles  with  a  Small  Part 85 

Quadrantal  Triangles    86 

CHAPTER  XI. 

Solution  of  Oblique  Spherical  Triangles 87 

Solution  by  Perpendicular  and  Napier's  Rules 88 

Solution  by  Cosine  Formula 90 

Solution  by  Napier's  Analogies 92 

Double   Solution    93 

Solution  by  Tangent  Formula  (the  Three  Sides  Given) 98 


STEREOGRAPHIC  PROJECTIONS. 

CHAPTER  XII. 

Definitions  and  Fundamental  Properties 100 

To  Project  a  Circle  with  Given  Pole  and  Polar  Distance 107 

Line  of  Centers  of  Circles  Through  a  Given  Point 109 

To  Draw  a  Great  Circle  at  a  Given  Angle  with  NS 110 

To  Find  the  Pole  of  a  Given  Circle Ill 

True  Length  of  a  Projected  Arc. Ill 

Terrestrial  Spherical  Triangles 112 

Great  Circle  Distances 112 

The  Celestial  Sphere 115 

Celestial   Coordinates    115 

The  Astronomical  Triangle 117 

Given  L,  d  and  h,  to  Project  Meridian  and  Horizon  and  Solve 121 

Given  t,  d  and  I/,  to  Project  on  Horizon  and  Meridian  and  Solve.  126 

Method  of  St.  Hilaire 130 

Given  t.  d,  h  to  Project  on  Equator  and  Solve 130 


CORRECTIONS  TO  LATEST  EDITION  OF  BROWN'S 
TRIGONOMETRY. 

(DECEMBER  10,  1916.) 

1.  Date  of  edition  not  correct. 

2.  p.  ix,  4th  line  from  bottom  of  page,  omit  the  words  "  Meridian 

and  Horizon." 

3.  p.  ix,  3d  line  from  bottom  of  page,  omit  the  words  "  on  Hori- 

zon and  Meridian/' 

4.  p.  6,  9th  line,  read"/!  ~~°Y  instead  of 

5.  p.  13,  6th  line,  omit  "  or  "  at  left-hand  edge  of  page. 

6.  p.  16,  next  to  last  line,  read  "  side  "  instead  of  "  sides." 

7.  p.   18,  3d  line,  read  "  sin  (180°  +  0)  =  -sin  0"  instead  of 

"sin  (180°+0)  =  -0." 

8.  p.  20,  last  line,  read  "cos"1  VT^r2"  instead  of  "cos^Vl-z2." 

9.  p.  21,  4th  line  from  bottom,  read  "  terms  of  sin  x  "  instead  of 

"  terms  sin  x."  _ 

10.  p.  26,  example,  middle  of  page,  read  "  5°  59'  50"  "  instead  of 

"5°  59'  #0"." 

11.  p.  28,  15th  line,  read  "  (68°  16')  "  for  "  (68°  15')." 

12.  p.  35,  9th  line  from  bottom,  read  "a  cotan  A"  for  "a  cotan  5." 

13.  p.  41,  5th  line,  read  "  Note  "  for  "  Notes." 

14.  p.  41,  2d  line  from  bottom,  read  "  135°  "  for  "  105°." 

15.  p.  47,  end  of  2d  line,  read  '7~  +  0Y  for  '7-~ 


16.  p.  48,  near  middle  of  page,  read  "  tan'1  "  for 


2  CORRECTIONS  TO  BROWN'S  TRIGONOMETRY. 

17.  p.  48,  next  line  of  work,  read 

"  6  -  ff  =  tan"1  x  -  tan-1  y  =  tan'1  %=*-  » 

1  +  xy 

18.  p.  50,  4th  line  from  bottom,  read  "  a  sin  B  "  for  "  a  sin  A" 

19.  p.  51,  in  Fig.  38,  read  "  D"  placed  directly  under  C,  for  "  <Z." 

20.  p.  51,  5th  line,  omit  "  and,"  reading  in  one  line, 

"in  EDO,  BD  =  aco$B=c  —  bco$A."  And  begin  next  line, 
"  Equating 

21.  p.  59,  5th  line,  read  "Arts.  59  and  60  "  for  "Arts.  48  and  49" 

22.  p.  61,  6th  line  from  bottom,  read  "(Fig.  42)"  for  "(Fig.  44) ." 

23.  p.  66,  last  line  of  Art.  77,  read  "  formula  "  for  "  formula." 

24.  p.  66,  last  line,  read  "  Art.  49  »  for  "  Art.  32." 

25.  p.  66,  next  to  last  line,  read  "79"  for  "75"  (number  of  Art.) . 

26.  p.  68,  3d  line,  read  "  35)  :  "  for  "  35)." 

27.  p.  69,  6th  line  from  bottom,  read 


28.  p.  70,  4th  line  from  bottom,  read  "  i  sin"  for 


29.  p.  75,  7th  line  from  bottom,  read  "  sin  b  "  for  "  sid  6." 

30.  p.  77,  2d  line,  read  "  cos2  $A  "  for  "  cas2  %A." 

31.  p.  78,  9th  line  from  bottom,  read  "(Art.  53)"  for  "(Art.  42)  ." 

32.  p.  86,  4th  line,  read  "  -45°  "  for  "  =45°." 

33.  p.  87,  5th  line  from  bottom,  read  "  A,  b,  c  or  A,  B,  c  "  for 

"  AbcoTABc" 

34.  p.  88,  6th  line,  read  "  solutions  "  for  "  solution." 

35.  p.  90,  16th  line,  read  "  and  </>  is  "  for  "  and  is." 

36.  p.  91,  6th  line  from  bottom,  read  "  JA  22°  40'  12"  "  for 

"A22°  50'  12"." 


CORRECTIONS  TO  BROWN'S  TRIGONOMETRY.  3 

37.  p.  92,  lower  right-hand  corner  of  page,  read  "  tan  9.79928  " 

for  "  tan  0.79928." 

38.  p.  94,  15th  line,  read  "6  =  93°  36'  2""  for  «  &  =  93°  35'  2"." 

39.  p.  98,  first  two  lines  of  Art.  115,  read  "This  problem"  for 

"  The  solution  of  this  problem." 

40.  p.  98,  3d  line  of  Art.  115,  read  "  (Art.  90)  "  for  "  (Art.  77) ." 

41.  p.  99,  10th  line  from  bottom,  read  "  cosec  0.06320  "  for 

"  cosec  0.36320." 

42.  p.   100,   1st  line  of  small  capitals,  read  "  ARTICLES  "  for 

"  CHAPTERS." 

43.  p.  103,  1st  line,  read  "  as  "  for  "  at." 

44.  p.  109,  2d  line  from  bottom,  read  "  point "  for  "  line." 

45.  p.  115,  8th  line,  read  "  A2  =  8h  3m  50s  E."  for  "  A^etc." 

46.  p.  115,  5th  line  of  Art.  132,  read  "  objects  "  for  "  subjects." 

47.  p.  118,  10th  line  from  bottom,  read  "  PZM  =  Azimuth  of  M  " 

for  "Plf^  =  etc." 

48.  p.  123,  3d  line,  read  "  PZ  =  9Q°  -L  "  for  "  PZ  +  900  -L" 

49.  p.  124,  last  line,  read  "  =70°  "  for  "  -70°." 

50.  p.  128,  1st  line,  read  "  ZM  "  for  "  Zm" 


PLANE  TRIGONOMETRY. 


CHAPTER  I. 

1.  Trigonometry  is  the  branch  of  mathematics  which  treats  of 
the  methods  of  subjecting  angles  and  triangles  to  numerical  com- 
putation. 

Plane  trigonometry  treats  of  plane  angles  and  triangles; 
spherical  trigonometry,  of  the  angles  and  triangles  formed  by  arcs 
of  great  circles  on  the  surface  of  a  sphere. 

2.  Comparison  of  Geometric  and  Trigonometric  Methods. — By 

the  methods  of  plane  geometry,  if  any  three  of  the  six  parts  of  a 
plane  triangle  are  given,  except  the  three  angles,  the  triangle  may 
be  constructed  and  the  unknown  parts  measured,  a  process  neces- 
sarily inconvenient  and  inaccurate. 

By  the  methods  of  trigonometry  the  unknown  parts  may  be 
found  by  numerical  computation  to  any  desired  degree  of  accu- 
racy; and  this  computation  is  called  the  solution  of  the  triangle. 

Before  such  a  solution  can  be  made  we  shall  have  to  express 
the  parts  in  numbers  by  the  adoption  of  suitable  units  of  meas- 
ure, and  define  certain  functions  of  angles  which  depend  upon 
their  magnitude. 

3.  Angles. — Although  the  solution  of  plane  triangles  involves 
the  consideration  of  angles  not  greater  than  180°,  many  problems 
of  mechanics  and  higher  mathematics  require  the  consideration 
of  angles  of  unlimited  magnitude. 

To  represent  such  an  angle,  we  consider  it  to  be  generated  by 
the  revolution  of  a  line  about  a  fixed  point.  Assume  such  a  line 

2 


2  PLANE  TRIGONOMETRY. 

to  start  in  coincidence  with  OA  and  to  revolve  into  the  position 
OB,  generating  the  angle  AOB,  in  the  direction  of  the  arrow 
(Fig.  1).  The  revolution  may  be  continued,  in  the  same  direc- 
tion, until  the  generating  line  comes  again  into  coincidence  with 
OB,  and  so  on  indefinitely,  the  angle  always  having  the  same 
boundary  lines,  but  with  each  revolution  greater  by  4,  8  or  any 
number,  4n,  of  right  angles.  The  side  OA  is  called  the  initial, 
and  OB  the  terminal,  side  of  the  angle. 


FIG.  1.  FIG.  2. 

Angles  generated  in  this  direction,  to  the  left,  or  counter- 
clockwise, are  usually  regarded  in  mathematical  works  as  positive, 
in  distinction  from  angles  formed  by  a  revolution  of  the  generat- 
ing line  in  the  opposite  direction,  to  the  right,  or  clockwise. 
Thus  the  angle  A  OB,  with  the  same  initial  and  terminal  sides,  but 
generated  in  the  direction  of  the  arrow  (Fig.  2),  is  regarded  as 
negative.  The  positive  direction  of  revolution  is,  however,  a  ques- 
tion of  convenience,  as,  for  instance,  in  the  determination  of  the 
errors  of  the  compass  in  ships,  where  angles  are  regarded  as 
positive  when  generated  clockwise. 

We  shall  always  regard  positive  angles,  unless  otherwise  stated, 
as  generated  in  the  counter-clockwise  direction. 

4.  Units  of  Measurement  of  Angles. — Angles  occurring  in  the 
solution  of  triangles  are  expressed  in  the  well-known  sexagesimal 
units  of  angular  measure :  degrees,  minutes  and  seconds — 


PLANE  TRIGONOMETRY.  3 

the  degree,  or  -fa  of  a  right  angle ; 
the  minute,  or  -fa  of  a  degree ; 
the  second,  or  -fa  of  a  minute — 

which  are  designated  by  the  symbols  1°,  1',  1",  respectively. 

5.  Measure  of  Arcs. — Since  arcs  of  a  circle  .are  proportional 
to  the  angles  they  subtend  at  the  center,  a  degree  of  arc  is  -g^-g- 
of  the  circumference,  and  is  subdivided  into  minutes  and  seconds 
just  as  the  degree  of  angle.    The  actual  length  of  a  degree  of  arc 

depends  upon  the  size  of  the  circle,  and  is  -^   r-  =  ^-^  in  a  circle 

obU         loU 

of  radius  r. 

6.  The  Circular  Unit. — For  certain  investigations  of  angles  and 
arcs  a  circular  unit  of  measure  is  used.     The  circular  unit  of 
arc  is  an  arc  of  the  same  length  as  the  radius ;  the  circular  unit  of 
angle,  called  the  radian,  is  the  angle  at  the  center  subtended  by 
this  arc. 

In  any  circle  the  semicircumference,  whose  length  is  TT  times 
the  radius,  subtends  an  angle  at  the  center  TT  times  as  great  as  a 
radian;  thus  TT  radians  and  180°  are  the  same,  and  the  equation 
connecting  the  two  units  is 

TT  radians =180°. 

In  writing  an  angle  in  radians  no  symbol  is  used ;  thus,  the  angle 
J  means  an  angle  £  of  a  radian;  the  angle  ~  means  ^-  or  1.5708 
radians,  the  equivalent  of  90°. 

7.  Table  for  Converting  Circular  into  Sexagesimal  Units: 

TT  radians  =       180° 

1       "      =         57.29578°+  Its  log  is  1.758123 

1       "      =     3437.7468'  "       3.536274 

1       "      =206264.8"  "       5.314425 


4  PLANE  TRIGONOMETRY. 

For  Converting  Sexagesimal  into  Circular  Units  : 

180°  =  3.14159265  radians  Its  log  is  0.497150 

1°  =  0.01745329       "  "       8.241877-10 

1'  =  0.00029089       "  "       6.463726  -  10 

1"  =  0.00000485       "  "       4.685575-10 

8.  In  numerical  work  the  minute  makes  a  convenient  unit;  and 
an  angle  6',  expressed  in  minutes,  is  changed  into  circular  units 
by  the  formula 

0=0.000290890', 

or  the  circular  value  of  1'  multiplied  by  the  number  of  minutes 
in  the  angle. 

For  the  reverse  process, 


Example  1:    Express  35°  42'  30"  in  circular  measure. 
Angle  0'  =  2142.5'  log     =3.33092 

log  1'  =  6.46373  -10 

0  =        0.62323  log     =  9.79465  -  10 

Example  2:    Express  0=  1.3786  in  angular  measure. 
0=       1.3786  log     =0.13944 

log  1'  =  6.46373  -10 

0'  =  4739.2'  log     =3.67571 

0=78°  59.2' 

Examples. 

1.  Give  the  value  in  angular  units  of 

7T  7T  2-JT  3-7T  3?T  9?T  17?T 

T'  T?  T~:  ~4~      .-8   J     4-J  ~13~' 

2.  Change  the  following  angles  into  circular  measure  : 

15°,  18°,  75°,  150°,  240°,  330°,  420°. 

3.  What  is  the  angle  at  the  center  of  a  circle  of  2  ft.  radius 
subtended  by  an  arc  12  inches  in  length  ?    Also  28  inches  ? 

Ans.     28°  38'  52V4;  66°  50'  42C'4. 


PLANE  TRIGONOMETRY. 


4.  Find  the  length  of  V  of  arc  on  the  earth's  equator,  assuming 
3962  miles  as  the  radius.    This  distance  is  the  geographical  mile; 
what  is  its  value  in  feet?  Ans.     6085.2  ft. 

5.  Find  the  length  in  feet  of  1"  of  arc  on  the  earth's  equator. 

Ans.     101.4  ft. 


FIG.  3. 

9.  Angular  Units  of  the  Mariner's  Compass. — Angles  indicated 
by  the  mariner's  compass  are  frequently  expressed  in  points,  each 
quadrant  being  divided  into  eight  points.  The  value  in  degrees 
of  one  point  is  therefore  11J°,  and  further  subdivisions  are  ex- 
pressed in  half-  and  quarter- points.  The  names  of  the  points  are 
indicated  in  Fig.  3. 


6  PLANE  TRIGONOMETRY. 

10.  Designation  of  Angles. — Angles  are  often  designated  by 
the  letters  of  the  Greek  alphabet,*  but  they  may  be  expressed  by 
any  other  algebraic  notation  according  to  convenience.  Their 
magnitude  may  be  expressed  numerically  in  either  unit,  although 
aliquot  parts  of  the  circumference  are  conveniently  expressed  in 
circular  measure,  as  2-n-  for  360°,  (IT— 6}  instead  of  (180°  — 0), 


7T 


9    for  90 


The  complement  of  an  angle  6  is  expressed  as  either  (90°  —  6)  or 


X 


The  supplement  of  the  same  angle  0  is  (ISO0  — 0)  or  (IT— 6). 

The  angular  space  about  the 
vertex    of    an    angle    may   be 
divided  into  four  quadrants  of 
1  90°  by  a  line  coinciding  with 

X  the  initial  side  of  the  angle, 

and  another  line  at  right 
angles,  as  in  Fig.  4,  the  angle 
XOY,  from  OX  to  OY,  being 
the  first  quadrant.  Angles  are 
PIG.  4.  said  to  be : 


In  the  first  quadrant,  when  less  than  90°. 
In  the  second  quadrant,  when  between  90°  and  180° 
In  the  third  quadrant,  when  between  180°  and  270° 
In  the  fourth  quadrant,  when  between  270°  and  360 


*For  convenience  the  Greek  Alphabet  is  given: 

A  a  Alpha  I     i  Iota  P   p  Rho 

B  j8  Beta  K    K  Kappa  2   a  Sigma 

T  7  Gamma  A    X  Lambda  T    T  Tau 

A  5  Delta  M  p  Mu  T    v  Upsilon 

E  e  Epsilon  N    v  Nu  $    0  Phi 

Z   f  Zeta  S    €  Xi  X  x  Chi 

H  tj  Eta  O    o   Oinicron  ^  i//  Psi 

0  6  Theta  II  it  Pi  ft   o    Omega 


PLANE  TRIGONOMETRY.  7 

It  is  important  to  note  that  this  designation  of  angles  refers 
JJonly  to  their  angular  magnitude. 

The  modern  navy  compass,  however,  is  graduated  in  degrees 
from  the  North  point  through  East,  South  and  West  from  0°  to 
360°.  Positive  angles  in  this  system  are  thus  generated  right- 
handed,  and  the  angular  magnitude  increases  through  the  four 
quadrants  in  the  opposite  direction  to  those  given  in  Fig.  4. 


s.i 


PLANE  TRIGONOMETRY. 


CHAPTER  II. 
THE  TRIGONOMETRIC  OR  CIRCULAR  FUNCTIONS. 

11.  Positive  and  Negative  Lines. — Let  the  vertex  of  an  angle, 
0,  be  taken  as  the  origin  of  a  system  of  rectangular  coordinates; 


X 


y' 


PIG.  7. 


y 

FIG.  8. 


the  initial  side  as  the  positive  axis  of  X;  and  the  terminal  side  of 
the  first  quadrant  as  the  positive  axis  of  Y.  Figs.  5,  6,  7  and  8 
show  the  angle  0  in  each  of  the  four  quadrants. 


PLANE  TRIGONOMETRY. 


9 


The  position  of  any  point  P  of  the  terminal  side  of  the  angle 
is  determined  by  its  perpendicular  distance  from  the  axes  X'X 
and  YY' ;  distances  from  X'X  are  taken  as  positive  when  P  is 
above  X'X,  negative  when  below ;  distances  from  the  axis  YY'  are 
considered  positive  when  P  is  to  the  right,  negative  when  to  the 
left. 

The  line  PM,  measured  parallel  to  YY',  is  the  ^-coordinate  of 
P,  or  simply  the  ordinate;  the  line  OM,  or  a  line  parallel  to  X'X 
from  P  to  YY',  is  the  ^-coordinate,  or  the  abscissa. 

The  terminal  side  of  the  angle,  OP,  is  positive  in  whatever 
quadrant,  and  PO  produced  back  through  the  vertex  is  negative; 
thus,  the  line  OP'  (Fig'.  5)  is  a  negative  line,  relatively  to  OP. 

12.  Definitions  of  the  Trigonometric  or  Circular  Functions  of 
Angles. — If  from  any  point  in  one  side  of  an  angle  a  perpendicular 
be  drawn  to  the  other  side,  produced  if  necessary,  a  right  triangle 
is  formed,  as  in  Fig.  5,  6,  7  and  8,  in  which  PM,  the  side  opposite 
the  angle,  is  the  perpendicular,  OM  the  base,  and  OP  the  hy- 
pothenuse. 

The  six  ratios  which  may  be  formed  between  the  three  sides  of 
this  triangle  are  called  the  trigonometric  ratios,  or  the  trigo- 
nometric or  circular  functions,  of  the  angle,  and  are  denned  as 
follows : 


The  sine  of  0 


The  cosine  of  6 


hypothenuse 


abbreviated  sin  9= 


hypothenuse   ' 
The  tangent  of  0      =  Pedicular  ,          «          tan  0= 


The  cosecant  of  6     = 


jr— r-,          "    *cosec0=^. 

perpendicular  y 

*  $ome  mathematicians  and  navigators  prefer  to  abbreviate  "  cose- 
cant "  as  "  esc,"  thus  making  formulas  symmetrical,  all  functions 
having  three  letters. 


10  PLANE  TRIGONOMETRY. 


The  secant  of  0         =    hyP°then^se      abbreviated,  sec  0=  -^ 

base  x 


The  cotangent  of  9  =  -  ,     ,          "          cot  «=  -  . 

perpendicular  y 

The  versine  of  0=  1  —  cos  0,  abbreviated  vers  0=   -     -  ,  is  fre- 

quently used  as  one  of  the  functions. 

It  is  important  to  remember  that  the  three  principal  functions 
are  the  sine,  cosine  and  tangent,  and  that  the  cosecant,  secant  and 
cotangent  are  the  reciprocals  of  the  first  three,  respectively. 

13.  These  functions  are  always  the  same  for  the  same  angle; 
for  if  the  perpendicular  be  drawn  from  any  other  point,  P1}  or  P1 
in  the  side  produced,  the  right  triangles  thus  formed  are  similar, 
and  the  ratios  of  the  homologous  sides  are  the  same,  not  only  in 
numerical  magnitude,  but  also  in  sign.     (Note  that  all  the  lines 
in  P'OM'  (Fig.  5)  are  negative,  in  accordance  with  the  assump- 
tions of  Art.  11.) 

The  ratios  are  in  general  different  for  different  angles  ;  for  two 
right  triangles  in  which  the  acute  angles  of  the  one  are  not  equal 
to  the  acute  angles  of  the  other  are  not  similar,  and  the  ratios 
of  their  homologous  sides  are  not  equal. 

The  ratios  thus  depend  on  the  magnitude  of  the  angle,  and  upon 
this  alone,  and  are  therefore  functions  of  the  angle. 

14.  Signs  and  Numerical  Value  of  the  Trigonometric  Func- 
tions for  Different  Quadrants.  —  Since  the  terminal  side,  r,  of  the 
angle  0  is  always  positive,  the  sign  of  the  trigonometric  function 
will  depend  only  on  the  sign  of  x  or  y,  or  both  x  and  y,  in  the 
triangle  of  reference  (Figs.  5,  6,  7  and  8). 

Thus  the  sine  is  positive  only  where  y  is  positive,  in  the  first  and 
second  quadrants  ;  the  cosine  is  positive  where  only  x  is  positive,  in 
the  first  and  fourth  quadrants  ;  the  tangent  will  be  positive  where 
both  x  and  y  have  like  signs,  in  the  first  and  third  quadrants. 


PLANE  TRIGONOMETRY. 


11 


From  these  simple  relations,  the  following  table  is  easily  made, 
showing  the  signs  of  the  functions  for  the  four  quadrants : 


Quadrant                   I 

II 

Ill 

IV 

Sine  and  cosecant 

+ 





Cosine  and  secant 

+ 

— 

— 

+ 

Tangent  and  cotangent 

+ 

— 

+ 

— 

Since  the  perpendicular  and  base  are  never  greater  than  the 
hypothenuse  of  the  right  triangle,  neither  the  sine  nor  the  cosine 
of  an  angle  is  ever  greater  than  unity;  the  cosecant  and  the  secant 
are  therefore  never  less  than  unity.  The  limits  of  the  numerical 
values  of  all  the  functions  are  given  in  the  following  table.  They 
may  be  readily  found  by  following  a  point  P  in  the  terminal  side 
as  it  revolves  around  0  as  a  center,  and  noting  the  signs  of  the 
coordinates  of  P,  and  the  value  of  the  ratio. 


6 

0  .. 

.90° 

90°... 

.180° 

180°. 

...270° 

270°. 

...360° 

Sine 

+0   .. 

•  +  1 

+  1   .. 

.+0 

—0 

..  .—1 

—1   . 

...—  0 

Cosine 

+  1   .. 

.+0 

—0  .. 

.—  1 

—1 

...—0 

+0   . 

...+1 

Tangent 

+0  .. 

•+« 

00   .  . 

.—0 

+0 

...+* 

GO   . 

...—  0 

Cosecant 

+«.. 

•  +  1 

+1  .. 

.+00 

GO 

.  ...—  1 

—1     . 

.  .  .  GO 

Secant 

+1  .. 

.+« 

QC   .  . 

.—1 

—1 

.  .  .  —  QO 

+00. 

...+1 

Cotangent 

+  05.. 

.+0 

—0   .  . 

.  00 

+  00 

....+0 

—0   . 

.  .  .  GO 

15.  Complementary  Functions. — Let  ABC  (Fig.  9)  be  a  right 
triangle,  with  the  right  angle  at  C.     Since 
Z?  =  90°— ,4,  we  may  deduce  from  this  rela- 
tion, and  the  definitions  of  the  trigonometric 
ratios,  the  functions  of  the  complement  of  an 

angle.    We  have : 

PIG.  9. 


12  PLANE  TRIGONOMETRY. 


=  —  :=cos£=cos(90°  -A), 


c 
tsMA  =  ~- 

cosecA  =  —  =  sec  B=  sec  (90°  -A). 
a 

On  account  of  this  relation,  the  sine  and  cosine,  tangent  and 
cotangent,  and  secant  and  cosecant  are  called  complementary 
functions. 

16.  Fundamental  Formulae.  —  In  any  of  the  right  triangles  of 
Figs.  5,  6,  7  and  8  we  have  always 


r2  r2 

From  the  definition  of  the  sine  and  cosine,  this  becomes 

sin2  0  +  cos2  (9=1.  (1) 

If  we  divide  the  same  equation,  x2  +  y2  =  r2,  by  x2,  we  obtain 

+  'x2  ~~  'x2  ' 
or 

l  +  tan20=sec20.  (2) 

If  we  divide  by  y2,  we  obtain 

l  +  cot20=cosec20.  (3) 

These  equations  are  graphically  represented  in  the  following 
figures,  the  designation  of  the  sides  of  the  right  triangle  being  so 
made  that  the  ratio  involved  in  the  definition  gives  at  once  the 

function ;  thus,  sin  6—  S11^     . 


PLANE  TRIGONOMETRY. 


13 


sin 


tan 


j 
FIG.  10. 


cor 


From  the  definitions  of  the  functions  we  have  the  following 
important  equations: 

111 


cosec  6  =  -. — -  , 
sin  6 

Since  we  always  have 


cot  6  = 


tan  6 ' 


sin  6  _  y  _._  x_  _    y 
cos  6        r        r  ""  a; 


tan  = 


sin  0 
cos  6 


•(4) 


17.  The  values  of  all  the  functions  may  be  expressed  in  terms 
of  any  one  function  by  means  of  the  formulae  of  Art.  16.  For  in- 
stance, 

sin  0=sin  9, 
cos0=  Yl  —  sin20, 

sin  0  sin  0 


tan0= 


cos  6       VI -sin2  (9 


cos0~  Vl-sin20 


,  etc. 


The  same  result  may  always  be  more  readily  obtained  by  mak- 
ing the  given  function  the  appropriate  side  of  a  right  triangle. 
Thus,  for  the  preceding  example,  the  right  triangle  is  given  in 

Fig.  11,  where  by  definition  sin  0=  — - —  .    The  value  of  any  func- 
tion in  terms  of  sin  6  is  then  derived  directly  from  its  definition. 


14 


PLANE  TRIGONOMETRY. 


FIG.  11. 


4 
FIG.  12. 


If  the  numerical  value  of  one  function  is  given,  the  graphic 
method  furnishes  the  simplest  means  of  finding  the  numerical 
value  of  all  the  other  functions.  As  an  example,  if  the  given 
function  is  tan  0=},  the  right  triangle  whose  perpendicular  is  3 
and  whose  base  is  4  (Fig.  12)  gives: 

tan  0= J,     sin  0=f,     cos  0=f,  etc. 

18.  Examples. 

1.  Find  the  remaining  functions  from  the  following  data  (Arts. 
16-17)  : 
(a)     sina=T57.          (b)     seca=-i-J-. 


(d) 


(b) 

(e) 


m*  —  n' 


(c) 

(f) 


COS  a  = 


COSa=  — 


2.  Is  the  equation  sec2  B— 


a  possible  equation  ? 

from  the 


3.  Derive  the  values  of  all  the  functions  of  45° 
isosceles  right  triangle  whose  two  equal  sides  are  a. 

Ans'.     sin  45°  =  -i=- ,  tan  45°  =  1. 

4.  Derive  all  the  functions  of  30°  and  60°  by  dividing  the 
equilateral  triangle,  of  side  a,  into  two  equal  right  triangles. 

Ans.     sin  30°  =4,  cos  30°=iV3,  tan  30°=£\/3. 

5.  Prove  the  following  identities : 

(a)  (tan0  +  cot0)sin0cos0=l. 

(b)  (sin0+cos0)2-f  (sin  0-cos0)2  =  2. 
(e)    (tan20-sin20)=tan20sin20. 

(d)   sec2  0+csc2  0-sec2  0  esc2  0. 

rl  — sin  0  _  1  — sin  0  _     cos  0 

COS0 


/ 
V 


PLANE  TRIGONOMETRY. 


15 


6.  From  the  relations  of  complementary  functions  find  a  value 
of  the  angle  involved  in  the  following  equations : 

(a)  sina=cos2a  (note  that  sinea  =  sin  (90°  — 2a),  and  thus 

a  =  90°-2a,  a  =  30°. 

(b)  tan2a=cot3a. 

(c)  CSC  I  a  =  SCO  a. 

(d)  sin  60  =  cos  39. 

(e)  sin(0-0)=J,  cos(0+<j&)=0. 

19.  Trigonometric  Functions  Represented  in  Numerical  Value 
by  the  Length  of  Lines. — Since  the  numerical  values  of  the  trigo- 
nometric ratios  are  independent  of  the  length  of  the  terminal  side, 


FIG.  14. 


we  may  suppose  the  point  P  to  generate  in  its  revolution  a  circle 
of  radius  unity,  as  in  Fig.  13.  Then,  from  the  figure,  x  and  y  are 
always  \ess  than  unity ;  and 


~ 


cos  0=  T  , 

or 

i  ' 


and  the  lengths  of  these  lines  represent  graphically  the  numerical 
values  of  the  functions  in  terms  of  r=l.  And  since  the  angle 
POY=9Q°  -6,  by  construction 

=  cot  6,  and  OT'  =  cosec  6. 


16 


PLANE  TRIGONOMETRY. 


When  the  angle  is  in  the  second  quadrant  (Fig.  14),  the  termi- 
nal side,  produced,  meets  the  tangent  at  A,  produced,  in  T  and  the 
values  of  the  secant  and  tangent  are  negative  ;  i.  e., 

sec  B-  OT,         tan  6=  AT; 


The  construction  of  the  functions  for  the  other  two  quadrants 
is  readily  made  (Figs.  15  and  16). 


PIG.  15. 

3d  Quadrant. 
AT  =tan  0,  positive. 
OT  =  sec  0,  negative. 
Fr'=cot0,  positive. 

=  cosec  6,  negative. 


20.  Reduction  of  the  Functions  of  Any  Angle  to  the  First 
Quadrant. — The  functions  of  any  angle  may  be  expressed  in  terms 
of  functions  of  an  angle  in  the  first  quadrant.  The  necessary 
equations  are  easily  derived  by  means  of  the  coordinates  of  a  point 
in  the  terminal  sidefc  of  the  angle,  as  used  in  the  general  definition 
of  the  functions. 


FIG.  16. 

4th  Quadrant. 
AT  =  tan  0,  negative. 
OT  =sec  6,  positive. 
YT'  =  cot  0,  negative. 
6,  negative. 


PLANE  TRIGONOMETRY. 


17 


21.  Functions  of  (90°  +  <9).—  Let 
X0P=90°  +  0  be  an  angle  in  the 
second  quadrant,  and  draw  OP'  per- 
pendicular to  OP,  thus  making  the 
angle  XOP'  =  B  (Fig.  17).  The  tri- 
angles P'OM'  and  POM  are  equal, 
since  the  angle  POX'  =  90°  -0;  and 
hence  y  =  x'  and  x=  —  y'. 

We  have,  from  the  figure  : 

sin  (90 


cosec(90 


COS  i 


—  =     cos  0, 


sec  (90°  +  0)  =  -^-=  -£-p 

" 

tan(90°  +  0)  =  £  =  — ^ 


-     sec0; 


=  —sin 


=  —  cosec  0 ; 


=  —cot 


cot(90°+0)=-  = F—  = 

i/  a; 


tan0. 


Expressed  in  words,  these  formulae  state  that  the  function  of  an 
angle  in  the  second  quadrant  is  the  complementary  function  of  its 
excess  above  90°  ,  with  the  proper  sign  of  the  desired  function  for 
the  second  quadrant. 

22.  Functions  of  (180°  -0).—  If  X'OP  (Fig.  18)  is  laid  off 
equal  to  0,  an  angle  XOP'  in  the  first  quadrant,  then  XOP=  180° 
—  6  ;  hence,  from  the  figure  : 

sin(180°-0)  =      sin0,     cosec  (  180°  -B)=     cosec  6; 

cos(180°  -6)  =  -  cos  0,        sec(180°-0)  =  -sec  0; 

tan(180°-0)  =  -tan  0,        cot(180°-0)  =  -cot  0. 

3 


18 


PLANE  TRIGONOMETRY. 


P' 


FIG.  19. 


23.  Functions  of  (180°  +0).—  Let  XOP  be  an  angle  0  (Fig. 

19),  and  XOP  '=  (180°  +0)  ;  then  by  definition  : 

—  <9 


tan  (180°  -Ml)  =     tanfl. 
The  value  and  signs  of  the  reciprocal  functions  are  obvious. 


24.  Functions  of  (270°db0).— Let  TOP  and  TOP  (Fig.  20) 
each  equal  an  angle  0  in  the  first  quadrant ;  then 


PLANE  TRIGONOMETRY.  19 


whence,  by  definition : 

sin  (270° -0)=      sin  (270°  +0)=  -£- = -^  =  -  cos  0, 

cos (270°  -0)  =  -  cos (270°  +  0)  =  ^  =  -^  =  -sin  0, 

7/  —  T 

tan  (270°  — 0)  =  —  tan  (270°  + 0)  = -*•?-=  —  =  +cot  0, 

x         —y 

the  other  functions  being  obvious. 

25.  Functions  of  (360° -0)  or  (-0).— In  Fig.  21  let  XOP 
and  XOP'  be  equal  angles,  0,  measured  from  the  initial  side,  OX, 
in  opposite  directions.    Then  XOP'  may  be  regarded  either  as  a 
negative  angle   ( —  0)   or  as  an  angle  in  the '  fourth  quadrant 
(360°  —  0),  and  we  have,  in  either  case, 

sin(-0)  =  sin(360°-0)  =  -sin  0, 
cos(  -0)  =  cos(360°  -0)  =  +cos  0, 
tan( -0)  =tan(360° -0)  =  -tan  0. 

26.  All  the  relations  of  functions  of  angles  in  the  various  quad- 
rants to  the  functions  of  an  angle  in  the  first  quadrant  may  be 
summarized  in  the  two  formulae 

f(n~±6)  =±co-f(0),  when  n  is  odd; 
and 


f(nr±6)  =  ±f(0),  when  n  is  even, 

to 

where  /(0)  is  any  desired  function  of  0,  and  co-/(0)  the  comple- 
mentary function  of  0,  with  the  appropriate  sign  of  the  function 


for  the  quadrant  of  (^-  ±  0)  . 


20  PLANE  TRIGONOMETRY. 

27.  Periodicity  of  the  Trigonometric   Functions. — We  have 
seen,  in  Art.  3,  that  the  initial  and  terminal  sides  of  a  given  angle 
may  indicate  an  angle  0,  or  the  same  angle  plus  one  or  more  revo- 
lutions of  the  terminal  side  through  360°,  or  2?r.    Hence,  from 
the  definitions  of  the  functions  of  angles,  it  is  evident  that  they 
remain  the  same  when  the  angle  is  increased  or  diminished  by  any 
multiple  of  360°,  or,  expressed  as  an  equation, 

sin  ( 2/iTT + 0)  =  sin  0,     cos  ( 2mr +  0)=cos6, 

and,  generally, 

f  (%**+$)  =f  (6), 

where  /  denotes  a  circular  function. 

Exercises. 

1.  Make  a  table  of  the  value  and  sign  of  all  the  functions  of 
0°,  30°,  60°,  90°,  120°,  etc.,  to  360°. 

2.  Make  a  table  of  the  functions  of  0°,  45°,  90°,  135°,  etc.,  to 
360°. 

28.  Inverse  Functions. — In  the  equation  z  =  sin  6,  x  is  said  to 
be  an  explicit  function  of  0.    To  express  6  as  an  explicit  function 
of  x,  the  equation  is  written 

which  is  read  "  6  is  the  angle  or  arc  whose  sine  is  x  "  and  sin'1  x 
is  called  an  inverse  function  of  x. 

To  express  9  by  any  other  inverse  function  most  conveniently, 
we  may  graphically  represent  9  and  its  given  function  as  parts  of 

a  right  triangle,  as  in  Fig.  22.    From  the  parts 

of  the  right  triangle, 


PLANE  TRIGONOMETRY.  21 

29.  Angles  Corresponding  to  Given  Functions.  —  We  have  seen 
that  for  a  given  angle  there  is  but  one  value  to  each  of  its  func- 
tions ;  but  if  we  have  to  find  the  angle  corresponding  to  a  given 
function,  there  is  an  indefinite  number  of  angles  which  have  the 
same  function. 

If  we  have,  corresponding  to  a  given  sine,  ~b,  an  angle  (3,  then  we 
have,  also  (from  Art.  22),  TT—  £,  and  (from  Art.  27)  an  indefinite 
number  of  angles  formed  by  adding  multiples  of  2-n-  to  these;  i.  e., 

A     27T+/J,     47T+  p,  etc.; 
TT-/J,     3*—&     57r-/?,etc., 

which  may  be  reduced  to  the  formula 

-l)n  p. 


Corresponding  to  a  given  cosine,  b,  if  ft  be  one  value,  then  — 
will  also  be  another,  since  cos  (3  =  cos  (  —  /?),  and,  generally, 

-1 


cos-     =  nir± 

Corresponding  to  a  given  tangent,  6,  if  /?  be  one  value,  then 
TT+/J  is  another,  since  tan  /?=:tan(7r+/?),  and,  generally, 

tan-1  &  = 


Equations  frequently  occur  which  are  expressed  in  terms  of  the 
functions  of  an  angle.  They  are  solved  by  the  elementary  proc- 
esses of  algebra,  after  first  transforming  the  equation  into  one 
which  contains  only  a  single  trigonometric  function. 

As  an  illustration,  let  it  be  required  to  solve  the  equation  sin  x 
=  tan2  x.  u 

Transforming  to  terms.sin  x,  it  becomes 


DUA    .*/  _  ; ^  , 

cos2  a;       1  —  sm2  x 
which  reduces  to 

sina;(l  —  sin  a;  —  sin2  x)  =  0. 


22  PLANE  TRIGONOMETRY. 

The  factor  sin  x=Q  gives 

X=UTT(\.  e.,  0,  TT,  2^r,  etc.). 

The  other  factor, 

sin  2  a;+sin  a;— 1  =  0, 

solved  as  a  quadratic  in  terms  of  sin  x  as  the  unknown,  gives 


sina;  =  ~l±v5  =0.61803  or  -1.61803. 
2 

Since  sin  x  cannot  be  numerically  greater  than  unity,  the  second 
value  is  obviously  impossible,  although  it  satisfies  the  original 
equation. 


30.  Examples. 

1.  Find  all  the  angles  less  than  2-ir  corresponding  to  the  fol- 
lowing functions : 

sin-1^    tan^VS,     co^TTp    cot'^-l),     sec-^-D. 

2.  What  is  the  sin  (cos-1 4)  ?  tan  (cot'1  x)  ? 

3.  Solve  the  following  equations : 

(a)  2  cos  0= sec  0.  Ans.     ^TT±  -j-. 

(b)  4  sin  ^=3  esc  (9.  Ans.     n7r±47r. 

(c)  3  tana;  — cot  a:  =0. 

Ans.     n7r±~>  (30°,  150°,  210°,  330°,  etc.) . 

(d)  tan0+cot0=2. 

(e)  3  tan2  0-4  sin2  0=1.  Ans.     sin  0=±—7~. 

V-4 

(f)  2sin0-tan0=0. 


PLANE  TRIGONOMETRY.  23 


CHAPTER  III. 
TRIGONOMETRIC  TABLES. 

31.  Tables   of  Natural   Sines,   Cosines   and  Tangents. — The 

numerical  values  of  sines,  cosines  and  tangents,  usually  called  the 
natural  functions,  have  been  computed  and  collected  in  many 
different  tables,  distinguished  by  names  indicating  the  degree  of 
approximation  to  which  the  values  are  given ;  thus  there  are  four-, 
five-,  six-  or  seven-place  tables,  giving  the  values  of  the  functions 
correct  to  that  number  of  decimal  places. 

In  Bow  ditch's  Useful  Tables,  Table  41  gives  only  the  natural 
sines  and  cosines  to  five  places,  for  each  minute  from  0°  to  90°. 
The  functions  of  intermediate  values  may  be  found  by  inter- 
polating for  the  seconds,  on  the  assumption  that  the  algebraic 
increase  of  the  function  is  proportional  to  the  increase  in  the 
angle.  Thus,  for  example,  if  we  wish  the  sine  of  30°  2 6'  35"  the 
sine  of  30°  26'  is  0.50654,  and  the  tabular  difference  for  one 
minute  between  26'  and  27',  is  .00025,  and  for  35",  ff  of  .00025 
or  .00015  and  the  interpolated  value  is  0.50669. 

Auxiliary  tables  of  proportional  parts  are  given  in  the  first  and 
last  column  of  each  page,  marked  "  Prop,  parts  25  " — in  the  ex- 
ample cited  above,  the  number  25  being  the  tabular  difference  for 
1'.  The  interpolation  for  the  35"  is  found  in  this  column  opposite 
the  35  in  the  column  of  minutes  immediately  adjoining. 

The  cosine  of  the  same  angle  30°  26'  is  0.86222,  the  tabular 
difference  for  1'  is  — 15,  and  the  interpolation  for  35"  is  f-§  X  — 15 
=  —  9;  the  interpolated  value  is  thus  0.86213,  since  the  cosine  is 
a  decreasing  function. 

The  auxiliary  table  for  this  interpolation  is  found  in  the  last 
column,  marked  "  Prop,  parts  16,"  and  the  proportional  part  for 


24  PLANE  TRIGONOMETRY. 

35",  opposite  the  35  in  the  column  of  minutes  immediately  ad- 
joining, is  9. 

The  columns  of  minutes,  marked  "  M,"  thus  serve  the  double 
purpose  of  giving  the  minutes  of  the  tabulated  angle  and  also  in- 
dicating the  seconds  for  which  the  value  of  the  "proportional 
part "  is  given  in  the  same  line. 

The  tabulated  "  proportional  part  "  for  a  given  page  is  made  on 
the  assumption  that  the  tabular  difference  for  1'  is  constant  for  the 
page ;  while  the  actual  difference  for  this  page  varies  from  26  to 
23.  When  greater  accuracy  is  desired,  the  actual  computation  is 
easily  made. 

Since  every  angle  between  0°  and  45°  is  the  complement  of 
another  angle  between  45°  and  90°,  every  function  of  an  angle 
less  than  45°  is  the  complementary  function  of  another  angle 
greater  than  45°.  Therefore  it  is  necessary  to  extend  the  table 
only  to  45°,  the  degrees  at  the  top  of  the  page  extending  from  0°  to 
45°,  those  at  the  bottom  from  45°  to  90°.  Each  number,  there- 
fore, in  each  column,  stands  for  the  function  named  at  the  top  of 
the  column  and  the  complemental  function  at  the  bottom  of  the 
same  column.  Thus,  in  looking  for  a  function  of  an  angle  less 
than  45°,  the  degrees  are  found  at  the  top  of  the  page,  and  the 
minutes  in  the  left-hand  column,  marked  "  M  " ;  for  an  angle 
greater  than  45°,  the  degrees  are  found  at  the  bottom,  and  the 
minutes  in  the  right-hand  column,  marked  "  M." 

32.  Tables  of  Logarithmic  Functions. — The  logarithms  of  the 
trigonometric  functions,  which  are  of  far  greater  use  in  computa- 
tions, are  tabulated  on  the  same  general  plan  as  the  natural  func- 
tions ;  and  the  tables  are  likewise  distinguished  by  the  number  of 
decimal  places  to  which  the  mantissas  are  carried. 

Table  44,  of  Bowditch's  Useful  Tables,  gives  the  logarithms  of 
all  six  functions  to  five  places.  Since  the  sine  and  cosine  of  all 
angles  and  the  tangents  of  angles  less  than  45°  are  less  than  unity, 


PLANE  TRIGONOMETRY.  25 

their  logarithms  are  negative.  To  avoid  the  use  of  such  loga- 
rithms, the  tabular  logarithms  of  all  the  functions  of  Table  44  are 
increased  by  10,  which  must  be  subtracted  from  the  tabular  loga- 
rithm to  give  the  correct  logarithm  for  use  in  computing.  For 
instance,  the  tabular  log  sin  30°  is  9.69897,  from  which  10  must 
be  subtracted,  making  the  correct  log  sin  30°,  9.69897  — 10. 

The  arrangement  of  the  degrees  from  0°  to  45°,  and  the  corre- 
sponding log  function,  at  the  top  of  the  page,  and  from  45°  to  90° 
at  the  bottom,  is  the  same  as  for  Table  41 ;  but  in  addition  there 
is  also  given,  at  the  opposite  end  of  each  column  of  minutes,  an 
angle  90°  greater. 

Thus  the  log  functions  of  any  angle  from  0°  to  180°  may  be 
taken  directly  from  Table  44,  without  resorting  to  the  use  of 
complemental  functions,  as  explained  in  Arts.  21  to  28.  An  in- 
spection of  the  table  shows  that  in  all  cases  the  minutes  of  a  given 
angle  are  found'  in  the  "  M "  column  immediately  below  the 
degrees,  when  these  are  at  the  top,  and  immediately  above,  when 
at  the  bottom,  of  the  page. 

33.  Tables  of  Proportional  Parts,  Table  44. — For  angles  inter- 
mediate between  the  minutes  of  the  table,  the  interpolation  for  the 
given  seconds  may  be  made  on  the  usual  assumption,  that  the 
change  in  the  logarithm  is  directly  proportional  to  the  change  in 
the  angle. 

For  the  five  degrees  at  each  end  of  the  first  and  second  quad- 
rants, the  tabular  difference  for  1'  is  given  for  each  function  in 
the  column  marked  "  Diff.  1' " ;  and  for  angles  found  on  these 
pages  the  interpolation  for  the  seconds  must  be  computed. 

It  should  be  noted,  however,  since  the  sine  and  cosecant,  tangent 
and  cotangent,  cosine  and  secant,  are  respectively  reciprocal  func- 
tions, that  the  log  of  one  of  these  pairs  is  the  colog  of  the  other, 
and  their  tabular  difference  is  the  same  numerically,  but  opposite 
in  sign. 


26  PLANE  TRIGONOMETRY. 

For  angles  outside  of  these  five-degree  limits,  the  proportional 
part  for  each  second  of  angle  is  contained  in  the  column  marked 
"  Diff./'  and  the  corresponding  seconds  are  always  found  in  the 
left-hand  column  of  minutes,  no  matter  where  the  degrees  of  the 
angle  may  be  found. 

These  auxiliary  tables  for  interpolation  depend  upon  the  as- 
sumption that  the  tabular  difference  for  1'  for  a  given  function 
is  the  same  for  the  whole  column.  For  angles  near  5°  or  85°,  this 
assumption  leads  to  an  error  which  may  be  as  great  as  12  units  in 
the  last  figures  of  the  logarithm.  It  is  therefore  advisable  to  com- 
pute the  interpolation,  when  the  tabular  differences  are  large. 
As  an  example  of  this  error,  the  log  sine  of  5°  59'  50"  is  found 
below : 

5°  59'  log  sin  =     9.01803-10 

Tab.  Diff.  for  50"  =  +        110 

5°  59'  Jb"  log  sin  =     9.01913-10 

The  tabular  difference  for  1'  between  59'  and  60'  is  120,  and  the 
proportional  part  for  50"  is  120  xf  =  100;  and  the  more  nearly 
correct  logarithm  is  9.09103  - 10. 

If  the  interpolation  is  made  backwards,  however,  for  10"  from 
5°  60',  the  log  sine  is  9.01901  —  10,  a  much  closer  approximation 
than  the  first. 

Example  1 :    Find  the  log  sine  and  log  cosine  of  28°  59'  24". 

28°  59'  sin      9.68534 — 10  cos     9.94189—10 

"  Diff."  2T,  "A"          +  9          «  Diff."  «  C  "      —  3 

28°  59'  24"  sin      9.68543—10  cos     9.94186—10 

The  tabular  difference  for  1'  at  59'  is  23,  and  the  interpolation 
for  24"  is  0.4x23  =  9.2,  or,  to  the  nearest  fifth  place,  9,  which  is 
found  in  the  column  marked  "Diff.,"  opposite  24  in  the  first 
column. 


PLANE  TRIGONOMETRY.  27 

Example  2:  Find  the  log  tangent  and  log  cosecant  of  59°  54' 
48". 

At  the  bottom  of  the  page  for  59°  (opposite  54'  in  the  right- 
hand  column),  is  found 

59°  54'  tan     0.23681  cosec     0.06291 

"Diff."  48",  "B"       +          23  "Diff.""C"  6 

59°  54'  48"  tan     0.23704  cosec     0.06285 

The  interpolation  for  48"  is  taken  from  the  column  marked 
"  Diff.,"  between  the  tangent  and  cotangent  and  opposite  48"  in 
the  left-hand  column. 

Example  3:  Find  the  log  cosine  and  log  tangent  of  111°  25' 
35".  The  degrees  111°  are  found  at  the  bottom  of  the  page  under 
the  first  column,  and  the  minutes  must  be  taken  in  the  same 
column ;  thus : 

111°  25'  cos    9.56247— 10n  tan     0.40646w 

"Diff."  35",  "A"        +          19  «Diff.""B"  —        22 


111°  25'  35"  cos    9.56266—  Wn  tan     0.40624n 

Note  that  the  minutes  increase  from  the  top  of  the  page,  and 
the  direction  in  which  the  interpolation  is  to  be  applied  is  found 
by  noting  whether  the  log  function  for  the  next  higher  minute 
is  greater  or  less.  As  both  the  required  functions  in  the  second 
quadrant  are  negative,  this  is  indicated  by  the  "  n  "  affixed  to  the 
logarithms. 

34.  The  Angle  Corresponding  to   a  Given  log  Function. — 

Except  for  the  sine  and  cosecant,  the  sign  of  the  function  indi- 
cates whether  the  corresponding  angle  is  in  the  first  or  second 
quadrant.  If  the  given  function  is  found  in  the  tables,  take  out 
the  degree  and  minute  corresponding  to  it.  Otherwise,  find  the 
two  tabular  functions  between  which  the  given  function  lies,  and 
note  the  difference  between  the  given  function  and  the  tabular 
function  corresponding  to  the  smaller  minute;  find  the  number  of 


28  PLANE  TRIGONOMETRY. 

seconds  in  the  column  "  M  "  on  tine  left  side  of  page,  corresponding 
to  this  difference,  and  add  to  the  degrees  and  minutes  already 
written  down.  When  greater  accuracy  is  required,  the  seconds 
may  be  computed  from  the  proportion  which  the  difference  bears 
to  the  tabular  difference  for  1'  (or  60"). 

If  the  function  is  the  sine  or  cosecant,  there  are  always  two 
corresponding  angles,  one  in  the  first  quadrant,  and  the  other  its 
supplement. 

Example  1 :    Find  the  angle  whose  log  cosine  is  9.56843  — 10. 

The  given  log  cosine  lies  between 

9.56854,  corresponding  to  68°  16', 
and 

9.56822,  corresponding  to  68°  17'; 

and  the  difference  between  it  and  the  tabular  function  for  the 
smaller  minute  (68°  Iff)  is  11,  corresponding  to  which,  in  the 
first  column,  "  M,"  is  20"  or  21". 

The  difference  for  11  corresponds'to  ^  x60  =  21";  and  the 
angle  is  68°  16'  21". 

Example  2 :    Find  the  angle  whose  log  tangent  is  n9. 62456  — 10. 

The  angle  is  in  the  second  quadrant ;  and  the  given  log  tangent 
lies  between 

9.62468-10,  for  157°  9', 
and 

9.62433-10,  for  157°  10'. 

Corresponding  to  the  difference  12,  in  "  B,"  we  find  20"  in  the 
left-hand  column,  "  M."  The  angle  is  157°  9'  20". 

35.  Functions  of  Small  Angles. — Let  6  (Fig.  23)  be  an  acute 
angle,  expressed  in  circular  units,  and  ABC  a  circular  arc  of 
radius  r,  the  point  C  determined  by  producing  the  perpendicular 
AM  to  C.  The  angle  AOC  then  equals  20.  Draw  the  two  tangents 
AT  and  CT. 


PLANE  TRIGONOMETRY. 


or 


or 


FIG.  23. 

From  plane  geometry, 

chord  AC<a,TcABC<AT+CT, 

2rsm0<2rO<2rtsin0, 
sin0<0<tan0, 


or 


sm  6 


(2) 


Now  let  the  angle  0  decrease  indefinitely,  approaching  zero  as  a 
limit ;  sec  6  approaches  unity,  and  when  0=  0,  sec  0=  1. 

n 

Hence  the  limit  of  - — z ,  which  lies  between  unity  and  sec  6, 
sin  6' 

is  unity,  or 


=  1. 


0=0 


30  PLANE  TRIGONOMETRY. 

Also 

tan  0      sin  0  /] 

— TT-  X  sec  0         = 


0 
When  0  is  a  small  angle,  we  may  then  write  the  equations 

sin  0=0,        tan  0=0 

as  approximate  formulae,  the  use  of  which  will  depend  upon  the 
degree  of  accuracy  required  in  a  given  problem. 
As  an  illustration, 

log  sin  1°  =  8.24186 -10, 
log  ^=8.24188 -10. 


The  tabular  difference  for  1'  for  1°  is  717 ;  hence  the  use  of  log  0 
for  log  sin  0  would  cause  an  error  in  an  angle  of  60'  of  TfT  X  60" 
=£".  For  2°, 

log  sin  =  8.54282 -10, 

log  ^=8.54291  -10. 

This  error  is  -gfa  x  60"  =  1.5". 

The  error  involved  will  be  shown  more  clearly  in  the  accurate 
expression  of  sin  0  in  terms  of  0. 

In  practice  with  five-place  tables,  the  formula 

sin0=0/sinl/ 
gives  the  last  place  of  five-figure  logarithms  correctly  to  23' ; 

tan  0=0' tan  1', 
to  only  16'. 

Example:    Find  the  log  sin  21'  21". 

21.35'  log  1.32940 

1'  log  sin  6.46373-10 

21.35'  log  sin  7.79313-10 


PLANE  TRIGONOMETRY.  31 

36.  Graphs  of  Trigonometric  Functions. — If  x  be  considered 
as  a  variable  angle  and  y  the  corresponding  trigonometric  func- 
tion, the  curves  of  the  trigonometric  functions  may  be  plotted  or 
traced  in  the  same  general  way  as  the  curves  of  algebraic  functions. 

In  order  that  the  linear  or  numerical  value  of  the  function  in 
terms  of  r=l  (Art.  19)  may  be  represented  by  the  graph,  the 
unit  of  values  must  be  the  same  for  the  angle  and  the  function ; 
hence  the  angle  must  be  expressed  in  circular  measure  in  terms 
of  7r=3.1416,  while  the  function  corresponding  to  the  given  angle 
is  expressed  in  terms  of  the  radius  unity. 

The  angle  increases  indefinitely,  while  the  functions  repeat 
themselves  in  the  same  order  from  every  multiple  of  £TT  or  360° ; 
it  follows  that  all  such  curves,  like  the  functions,  will  be  periodic. 
As  we  desire  to  show  principally  the  limiting  values  of  the  func- 
tions, we  shall  trace  the  curves  instead  of  plotting  them. 

In  the  curves  of  sines,  the  sinusoid, 

y  =  sinx,  (1) 

the  limiting  value  of  y  is  1  for  -^- ,  and  —  1  for  ~  ,  repeating  these 

values  in  succession  for  every  addition  of  2ir  to  these  angles. 

The  curve  crosses  the  axis  of  xf  i.  e.,  y  =  Q,  when  x  =  Q,  TT/^TT,  3ir 
and  so  on  indefinitely  for  every  multiple  of  TT. 

To  find  the  direction  of  the  curve  for  these  points,  we  have, 
from  (1)  and  Art.  35, 

y  _  sin  x~]       _  i 

x          x    J x=0  ~ 

To  find  the  direction  of  the  curve  at  X—TT,  when  y  =  0,  we  have 
to  resort  to  the  equation 

y=sin(7r+x)  =  —sin  x, 
or 

y_  _  sin  (IT  +  a;)  _    ^  sinafl       _  _  ^ 
x  x  x   \  x=0  ~ 


32 


PLANE  TRIGONOMETRY. 


Hence  the  curve  cuts  the  axis  of  x  at  the  angles  of  45°  and  135°, 
for  x—  0  and  X  =  TT  respectively,  and  is  traced  in  Fig.  24. 


0 


2.TT 


FIG.  24. 

The  curve  y  =  cosx,  since 

cos  #=sin  f~  —  x\, 

is  a  succession  of  similar  curves,  except  that  the  phase  is  advanced 
90°,  or^-  (i.  e.,  for  z-0,  y  =  l,  etc.). 

/v 

The  curve  y  =  tan  x  crosses  the  axis  of  x  at  the  origin,  at  an 
angle  of  45°,  since 

y_  _  tan  afl        _  -, 
~z~  "       *    J  ^=o  ~ 

When  a;  approaches  -^- ,  or  90°,  since  tan  $=s—    the  tangent, 
2  cos  a; 

or  #,  approaches  oo  as  a  limit ;  when  x  passes  through  ^ ,  the  cosine 

becomes  negative  and  the  ratio  approaches  —  oo  as  a  limit.    Hence 
the  value  of  the  ordinate  y  changes  from  +  oo  to  —  oo  as  x  passes 

through  the  value  -^-  into  the  second  quadrant. 


PLANE  TRIGONOMETRY. 


FIG.  25. 


Since  tan  (TT  +  X)  =tan  x,  the  direction  of  the  curve  in  passing 
through  TT  becomes 


JL  — 
x  ~ 


tana 


_  - 


and  the  curve  consists  of  a  series  of  similar  curves  always  crossing 
the  axis  of  x  at  an  angle  of  45°,  for  every  value  of  x  which  is  a 
multiple  of  ?r. 


The  curve  i  —  sec  x— 


cosx 


:    The  limiting  values  of  y  are  +1 


for  every  value  2mr  of  x,  —  1  for  every  value  (2n  —  1)  ?r  of  x;  and 
for  every  odd  multiple  of  -^-  the  value  of  y  changes  from  +  oo  to 

-  oo,  or  the  reverse,  as  x  passes  through  the  value  (2n  —  1)  -£•  . 


34 


PLANE  TRIGONOMETRY. 
Y    i 


PIG.  26. 

The  logarithmic  curve  y  =  logo  xt  on  account  of  its  showing  the 
limits  of  the  logarithms  of  the  trigonometric  function,  is  given 
below. 

[o=10]. 

y=     0        when  x—  1 
y      y=~l  "     x=.l 

y=-2  "     z=.01 

y=  —  oo  "     x—  0 

y=     1 


O 


x=  a 


Y' 


FIG.  27. 


PLANE  TRIGONOMETRY. 


35 


a 


CHAPTEE  IV. 
SOLUTION  OF  PLANE  RIGHT  TRIANGLES — POLAR  COORDINATES. 

37.  The  tabulated  values  of  the  trigonometric  functions  and 
their  logarithms  enable  us,  from  their  definitions  alone,  to  effect 
the  numerical  solution  of  right  triangles.  Omitting  the  right 
angle,  if  two  of  the  remaining  five  parts  of  the  triangle  are  given, 
one  of  them  being  a  side,  each  of  the  other  parts  may  be  found 
from  a  simple  equation  which  expresses  the  D 

definition  of  the  trigonometric  function  in- 
volved, in  terms  of  the  two  given  parts  and 
the  required  part. 

Let  ABC  (Fig.  28)  be  a  plane  right  tri- 
angle, C  the  right  angle,  and  a,  &,  and  c  the 
sides  opposite  the  angles  A,  B  and  C,  re- 
spectively. The  equations  necessary  for  the    **  6          ^ 
solution,  depending  on  the  parts  given,  are :                 FIG.  28. 

a= c  sin  A  =  c  cos  B=  I  tan  A  —  I  cotan  B=  Vc2  — 62, 
b  =  c  sin  B=  c  cos  A  —  a  tan  B= a  cotanf^—  Vc2  —  a2, 
c  =  a  cosec  A  =  a  sec  B  =  I  sec  A  =  b  cosec  B 

It  is  not  necessary  to  memorize  the  formulae,  as  they  are  readily 
derived  for  any  given  case  from  the  definitions  of  the  trigo- 
nometric ratios. 

It  is  always  advisable  to  make  a  rough  sketch  of  the  triangle, 
indicating  the  given  parts,  and  also  a  form  for  the  entire  computa- 
tion, before  taking  from  the  tables  any  of  the  desired  functions. 

Example:    Given  A  =  35°  16'  25",  c=672.34. 


36  PLANE  TRIGONOMETRY. 

The  formulae  are : 

B=9Q°—A,        a=csinA,         b  =  ccosA. 
The  solution  is : 

A  =  35°   16'  25"  sin  9.76153-10         cos  9.91190-10 

c  =  672.34  log  2.82759  log  2.82759 

a  =  388.25  log  2.58912 


5  =  548.90  log  2.73949 

When  there  is  no  liability  of  confusing  the  natural  and  loga- 
rithmic functions,  the  abbreviations  log  sin,  log  cos,  etc.,  are 
usually  still  further  abbreviated  into  sin,  cos,  etc.,  as  in  the  above 
example. 

38.  Examples. 

1.  A  =  44°  10'  38",  c  =  24.896. 

Ans.     a=  17.350,  6  =  17.854. 

2.  a=  396.25,  6  =  645.32.     Ans.     1  =  31°  33'  4",  c  =  757.25. 

3.  c  =  98.245,  a=95.573. 

Ans.     5=13°  23'  38",  6  =  22.757. 

4.  5  =  27°  9'  14",  a  =  35.421.        Ans.     6  =  18.168,  c  =  39.81. 

39.  The  examples  given  below  involve  the  solution  of  right  tri- 
angles, the  known  parts  of  which  are  found  by  the  measurement  of 
horizontal  distances  or  vertical  heights,  and  vertical  or  horizontal 
angles. 

A  vertical  line  is  a  line  determined  by  a  plumb-bob. 

A  vertical  plane  is  a  plane  determined  by  a  vertical  line. 

A  horizontal  plane  is  a  plane  perpendicular  to  a  vertical  line, 
and  for  small  distances  it  may  be  regarded  as  coinciding  with  a 
body  of  water  or  other  liquid  at  rest. 

An  angle  of  elevation  is  the  vertical  angle  of  an  object  above  a 
horizontal  plane. 

An  angle  of  depression  is  the  vertical  angle  below  a  horizontal 
plane. 


PLANE  TRIGONOMETRY. 


37 


40.  Examples. 

(To  be  worked  without  logarithms.) 

1.  At  a  distance  of  100  feet  from  the  foot  of  a  flag-staff,  the 
angle  of  elevation  of  its  top  is  30°;  what  is  the  height  of  the 
staff?  Ans.     57.735  ft. 

2.  The  angle  of  elevation  of  the  top  of  a  flag-staff  is  found  at 
a  certain  point  to  be  60° ;  150  feet  further  away,  in  the  same 
vertical  plane,  the  angle  is  found  to  be  30° ;  find  its  height  above 
the  horizontal  plane.  Ans.     129.9  ft. 

3.  The  angles  of  elevation  of  the  top  and  bottom  of  a  flag- 
staff 50  feet  high,  standing  on  the  top  of  a  building,  from  a  point 
in  the  plane  are  found  to  be  60°  and  45°  respectively;  what  are 
the  height  of  the  building  and  the  distance  of  the  point  of  ob- 
servation? Ans'  68.3  ft. 

4.  The  angle  of  elevation  of  a  balloon  bearing  N.W.  from  A  is 
45  ° ;  from  B  its  angle  of  elevation  is  the  same,  and  its  bearing 
N.E.;  A  and  B  are  one  mile  apart  on  a  line  running  E.  and  W. 
Find  the  height  of  the  balloon.  Ans.     3733  ft. 

5.  Later  the  balloon  is  due  north  of  A,  and  its  elevation  is  60°  ; 
at  the  same  instant  its  elevation  from  B  is  45°  ;  what  is  its  height? 

6.  A  ship  steaming  N.E.  by  E.  observes  a  lighthouse  bearing 
N".  by  E. ;  after  steaming  18  miles,  the  lighthouse  bears  N.W.  by 
N". ;  what  is  the  distance  of  the  light  at  the  time  of  the  last  bearing  ? 

7.  A  ship  steaming  S.W.  observes  a  lighthouse  bearing  W.S.W., 
and  after  steaming  16  miles  observes  the  light  bearing  W.    How 
much  further  will  she  steam,  on  the  same  course,  to  pass  the  light- 
house at  the  least  distance  ? 


41.  Projections. — The  orthogonal  projection,  P,  of  a  point  A, 
upon  a  given  line  OX,  is  determined  by  the  perpendicular  AP 


38 


PLANE  TRIGONOMETRY. 


P(r0> 


o 


upon  the  line  OX.  The  projection  of  the  line  AB  upon  the  given 
line  is  obviously  PQ;  its  projection  upon  a  line  OF  at  right  angles 
to  OX  is  RS.  The  projections  of  a  given  line  of  length  r  upon 
two  axes  at  right  angles  are  usually  called  the  horizontal  and 
vertical  projections  of  the  line,  and  from  the  right  triangle  in- 
volved in  the  construction  we  have  at  once,  calling  these  pro- 
jections x  and  y  respectively, 

x—r  cos0, 

y  —  r  sin  0, 
where  6  is  the  angle  of  the  line  with  OX 

42.  Polar  Coordinates. — Besides  the  familiar  method  of  fixing 

the  position  of  a  point  in  a  plane 
by  two  rectangular  coordinates,  we 
may  employ  the  method  of  polar 
coordinates  in  which  the  position 
of  the  point  is  determined  by  its 
distance  r=OP  from  a  fixed  point 
0  in  the  line  OA,  with  which  OP 
makes  the  angle  6,  as  shown  in 
Fig.  30.  The  point  0  is  called  the  pole  or  origin,  and  the  line  OA 
the  initial  line;  the  distance  r  is  the  radius  vector  and  6  the 
vectorial  angle  of  the  point  P,  designated  by  the  symbol  (r,  0). 
The  angle  0  is  measured  counter-clockwise  if  positive ;  clockwise  if 
negative ;  the  distance  r  is  measured  along  the  terminal  side,  of 
the  angle  0  if  r  is  positive,  in  the  opposite  direction  if  r  is  negative. 

43.  Transformation   of   Coordinates. — The   relation   between 

polar  and  rectangular  coordinates  is 
simple  where  they  have  the  same  origin 
and  the  initial  line  coincides  with  the 
re-axis,  as  shown  in  Fig.  31.  The  fol- 
x  lowing  equations  connecting  the  two 

systems  are  readily  derived  from  the 
FIG.  31.  figure: 


P(-r,0) 


FIG.  30. 


0 


PLANE  TRIGONOMETRY.  39 

x=r  cos  0,~] 


In  the  following  examples,  where  we  have  given  x  and  y  to  find  r 
and  By  we  may  regard  the  process  as  the  solution  of  two  simul- 
taneous trigonometric  equations 


rcos    —x, 
rsin  Q—, 


or  as  finding  the  polar  coordinates  of  a  given  point  from  its 
rectangular,  an  extension  of  the  ideas  involved  in  the  solution  of 
a  right  triangle  to  angles  of  any  magnitude. 
Dividing  the  given  equations,  we  have 


from  which  we  determine  0,  and  thence  get 
r=x  sec  Q—y  cosec  6. 
Example:    Given  x=  17.366,  y  —  20.462,  find  r  and  0. 

check 

#  =  29.462  log  1.46926  log  1.46926. 

x-  17.366  log  1.23970       log  1.23970. 

0=  59°  29.0'     I  tan  0.22956     I  sec  0.29432    I  esc  0.06475. 


r=34.199  log  1.53402       log  1.53401. 

Examples. 

Find  the  polar  or  spherical  coordinates  of  each  of  the  fol- 
lowing points,  giving  the  logarithmic  solution  and  check : 

(a)  (3,  4),  (b)  (-2.67,  3.48),  (c)  (2.72,  -6.76),  (d) 
(_3.478,  -7.286),  (e)  (*V3,  1),  (f)  (-2.7183,  -.36777), 
(g)  (6.765,0),  (h)  (0, -7.2143). 


40  PLANE  TRIGONOMETRY. 

44.  Equations  of  Curves  in  Polar  Coordinates. — The  equation 

of  a  curve  in  polar  coordinates  may  be 
found  by  expressing  its  geometric  defini- 
tion in  polar  coordinates.  For  instance, 
if  the  pole  0  is  on  the  circumfer- 
ence, and  the  initial  line  is  the  diam- 
eter of  the  circle  through  0,  as  in  Fig. 
32,  we  have  at  once,  since  APO  is  a 
right  angle, 
32.  r=2acos0 

as  the  polar  equation  of  the  circle  with  the  given  origin  and  initial 
line. 

The  polar  equation  thus  found  is  transformed  into  rectangular 
coordinates  by  substituting  the  values  of  r  and  cos  6  given  above, 

W~2_|_      2_Q       ^_    _  2flff 

which  reduces  to 

x2  +  y2  —  2ax. 

45.  Tracing  or  Plotting  a  Curve  in  Polar  Coordinates. — When 
the  equation  of  a  curve  is  given  in  polar  coordinates,  we  may  either 
plot  a  sufficient  number  of  points,  by  computing  the  value  of  r  for 
the  corresponding  angle  0  and  connecting  the  points  with  a  smooth 
curve,  or  we  may  trace  the  curve,  as  in  rectangular  coordinates, 
by  finding  its  limits  and  critical  points  and  sketching  in  the  curve. 
The  following  examples  illustrate  both  methods. 

Plot  the  locus  of  the  equation 

r  =  2a  cos  0 
by  computing  the  value  of  r  for  every  15°. 


PLANE  TRIGONOMETRY. 


41 


Taking  a  equal  to  one  inch,  the  values  of  r  are  found  from  a 
table  of  natural  trigonometric  functions  as  follows : 

0     0     15°      30°      45° 
r     2     1.93     1.73     1.41 


60°      75° 
1.00     0.52 


90' 
0 


N"ote^f  that  the  values  of  r  from  90°  to  180°  are  numerically  the 
same,  but  opposite  in  sign,  as  from  90°  to  0°  in  reverse  order.  The 


FIG.  33. 


first  half  of  the  curve  is  thus  plotted  in  Fig.  33,  and  the  smooth 
curve  drawn  in. 

Trace  the  curve  whose  polar  equation  is 

r=acos20. 

The  limits  of  r  are  seen  to  be  a  for  the  angle  0°  and  180°,  and 
-a  for  90°  and  27Q°.  The  curve  passes  through  the  pole  (r=0) 
for  values  of  45°,  1?5°,  225°,  and  315°,  making  the  same  angles 
in  each  case  with  the  initial  line  as  the  radius  vector. 


PLANE  TRIGONOMETRY. 


FIG.  34. 

The  form  at  the  pole  is  first  drawn  ;  then  starting  from  A  the 
curve  is  sketched  in  through  the  pole  to  Bf  from  B  through  the 
pole  to  G,  from  C  through  the  pole  to  D,  from  D  through  the  pole 
back  to  A. 

46.  Examples. 

1.  Trace  the  curves  given  by  the  following  polar  equations  : 

(a)  rsm6=a. 

(b)  rcos0=a. 

r=a(l  —  cos  0)   (cardioid). 


(c) 
(d) 
(e) 
(f) 


=  a2  cos  20(lemniscate). 


r=asin30. 

2.  Transform  the  following  curves  from  polar  to  rectangular 
coordinates  : 


(a) 

(b)  r=a. 

(c)  r=asecQ. 

(d)  r=2aBin 


PLANE  TRIGONOMETRY. 


43 


CHAPTEE  V. 

FUNCTIONS  OF  THE  SUM  OR  DIFFERENCE  OF  Two  OR  MORE 
ANGLES.    OF  MULTIPLE  ANGLES. 

47.  The  Sine  and  Cosine  of  the  Sum  or  Difference  of  Two 
Angles. — We  have  the  trigonometric  functions  of  two  angles, 
6  and  0',  and  wish  to  find  the  sine  and  cosine  of  (0+0')  and 
(0—0'),  in  terms  of  the  sine  and  cosine  of  0  and  0'. 

Let  POM  (Fig.  35)  be  the 
sum  of  the  two  given  angles. 
From  P,  any  point  in  the  side 
OP  (which  is  the  common 
terminal  side  of  (0+0')  and 
of  ff=NOP),  draw  PM  per- 
pendicular to  OM  and  PN  per- 
pendicular to  ON;  then,  from 
plane  geometry,  the  angle  NPM 
=  0.  Also  draw  NQ  perpen- 
dicular to  PM  and  designate  PIG  35 
the  magnitudes  of  the  various 

lines  as  in  the  figure.    From  the  definitions  of  the  trigonometric 
functions,  we  have 


sin(0+0')  = 


PM 


e± 
c 


OP          c      -  b       c 
=  sin  0  cos  tf  +  cos  0  sin  0' ; 


~d 


also 


OM 


a  — 
c 


"  OP 

=  cos  0  cos  0'  —  sin  0  sin  0 ; 


A    A     JL  .  A 

b       c     ~  d       c 


44  PLANE  TRIGONOMETRY. 

In  the  applications  of  the  general  definitions  of  the  trigo- 
nometric functions,  we  have  seen  that  the  definitions  apply  to 
angles  of  any  magnitude  if  we  consider  the  signs  of  the  auxiliary 
lines  involved  in  the  definitions.  It  follows  that  trigonometric 
formulae  deduced,  with  due  regard  to  signs,  from  geometric  figures 
are  of  general  application.  We  may  therefore  assume  that  ff  may 
be  replaced  by  ( —  6'),  in  the  above  formula?,  giving  thus  the  sine 
and  cosine  of  (6— 9')  (see  Art.  25)  : 

sin(0— ff)  =sin  0cos  &  —  cos  6  sin  0', 
cos(0  —  0')  =cos  6  cos  ff  -\-  sin  0  sin  6' '. 

These  four  formulae  are  fundamental  in  the  consideration  of 
functions  of  two  angles  and  should  be  carefully  memorized: 

sin(0+0')  =sin  6  cos  0'  +  cos  0  sin  0',  (5) 

cos(0+0')  =cos  0  cos  0'-sin  0sin  0',  (6) 

sin  (0-0')  =sin  0  cos  0' -  cos  0  sin  0',  (7) 

cos(0-0')  =cos  0  cos  0'  +  sin  0sin  0'.  (8) 

As  an  exercise,  prove  by  use  of  these  formulae  the  results  derived 
geometrically  in  Arts.  21-26;  thus,  from  (6), 

cos(90°+0)=cos90°  cos0-sin90°  sin  0= -sin  0. 

48.  Tangent  of  (0±0'). — Dividing  the  first  of  the  formulae  of 
Art.  47  by  the  second,  and  the  third  by  the  fourth,  we  get 

+      ta  \  a\—  sm  0  cos  0'4-cos  0  sin  0' 

V>  ~  eos  6  cos  ff- sm0  sin  ff  ' 
,      (fiff\_  sin  0  cos  6'  —  cos  0  sin  ff 
}~  cos  0  cos  0'  + sin  0  sin  0" 

If  each  term  in  each  of  these  fractions  be  divided  by  cos  0  cos  0', 


PLANE  TRIGONOMETRY.  45 

49.  Functions   of  Double   Angles.  —  If,   in  the   functions  of 
(0  +  0')  (Equations  5-10),  we  make  0=0',  there  results: 

sin  20  =2  sin  B  cos  0,  (11) 

cos  20=  cos2  B  -sin2  0,  (12) 


If  we  replace  cos2  6  by  1  —  sin2  6,  and  sin2  6  by  1  —  cos2  0,  in  the 
value  of  cos  20, 

cos  20  =1-2  sin2  0,  (14) 

cos  20=2  cos2  0-1.  (15) 

50.  Functions  of  the  Half  -angle.  —  Since  0  J-s  a  variable  angle, 
we  may  replace  20  by  <f>,  and  0  by  J<£,  in  formulae  of  Art.  49; 
getting  from  (14)  and  (15),  after  transposition, 

2  sin2  40  =  1  -cos  <£,  (16) 

2cos244>=l  +  cos</>,  (17) 

and,  from  (16)  -(17), 


_ 
2 

These  relations  are  three  of  the  most  useful  formulae  of  trigo- 
nometry. If  we  rationalize  first  the  numerator  and  then  the 
denominator,  (18)  is  freed  from  radicals: 


tan     -  <        _     -<  M9) 

2  " 


51.  Graphic  Derivation  of  Functions  of  the  Half  -angle.  —  Let 

0  be  an  angle  at  the  center  of  the  circle  ;  then  J0  is  the  inscribed 
angle  subtended  by  the  same  arc,  as  in  Fig.  36. 
Calling  the  radius  unity,  we  have  : 


46  PLANE  TRIGONOMETRY. 

From  these  data  and  the  definition  of  the  trigonometric  functions, 
we  derive: 


whence 


tan    e= 


sin  6 


1  +  COS0 

1  —  cos0 


1  — cos0 
sin  6     ' 


tan2  U= 


1  +  COS0  * 


FIG.  36. 


52.  Examples. 

1.  Find  the  value  of  sine,  cosine  and  tangent  of  15°  from  the 
functions  of  (45°  -30°). 

2.  Find  the  functions  of  15°  and  22J°  from  the  formula  of 
Art.  50. 

3.  Show  that  the  value  of  the  functions  of  15°  derived  in 
Example  2  are  the  same  as  those  derived  in  Example  1. 

4.  Let  &  —  ^Q  in  formulae   (5)   and   (6)   and  thus  derive  the 
sine  and  cosine  of  36. 

Ans.     sin  30  =  3  sin  6  —  4  sin3  0,  cos  30=4  cos3  6  —  3  cos  6. 


PLANE  TRIGONOMETRY.  47 


5.  From  Art.  50,  tan  --  =  cosec  <f>  —  cot  <£,  show  that 

T.~~ 

tan  f  ~  +  -£-  J  =  sec  B  +  tan  0.    Assume  <f>  =  (~  +  0  j  . 

6.  Show  that  sine  2x=  .  2  *anx    and  cos  2x=  ' 


.  2  . 

1  +  tan2  x  1  +  tan2  a; 

7.  Show  that  tan(45°  +  z)  =cot(45°  -a;)  =   l  +  i^x  . 

1  —  tan  x 

8.  From  formula  (18)  show  that 

I  it        x\         /I  —  sin  :r       1  —  sin  # 
tan   ——  -     zzA/i  —  bm  ^  —  -  -  =sec  a;  —  tana:. 

U        2y       Vl  +  sino;          cos  a: 

53.  Formulae  for  Transforming  Sums  into  Products.  —  From 
the  four  formulae  of  Art.  47,  we  get  by  addition  and  subtraction  : 

sin(0+0')  +sin(0-0/)  -  2  sin  6  cos  0', 
sin(0  +  0/)  -sin(0-0')  =  2  cos  0  sin  ff, 
006(0+?)  +cos((9-(9')  =  2  cos  0  cos  0', 
008(0+?)  -cos(0-^)  =  -2  sin  0  sin  0'. 

If  we  take 

0+0'  =  $, 

e-ef=<j>'f 

from  which 


and  substitute  these  values   in  the  preceding  equations,  they 
become  : 


sin<£  +  sin</>'=     2  sin  i(</>  +  <#>')  cos  £  (<£  —  <£'),  (20) 

sin^>  —  sin<£'=     2  cos  i(^>  +  ^>')  sin  |(<£  —  0'),  (21) 

cos  </>  +  cos  <£'=     2cos  J(<^>  +  ^')  cos  i(<^>  —  </>'),  (22) 
cos>  —  cos  <>'=—  2  si 


48  PLANE  TRIGONOMETRY. 

From  the  quotient  of  the  first  of  these  by  the  second,  and  of  the 
third  by  the  fourth,  we  obtain  : 

sin<4-sin<'       tan 


—  sin  ^  ~tan^(<£  —  <f>') 


54.  Sum  of  Two  Inverse  Functions.  —  If  we  have  two  angles 
given  as  inverse  functions,  the  formulae  of  Art.  48  may  be  em- 
ployed to  express  the  sum  or  difference  of  the  two  angles  as  inverse 
functions. 

If,  for  instance,  #  =  tan  6,  and  ?/  =  tan  &,  formulae  (9)  and  (10) 
become,  since  6—  tan'1  x  and  0'^tan'1  y, 


0+6'  =  tan'1  x  +  tan"1  y  =  tan 

l-xy 

and 

0-j-0'  =  tan"1  x  —  tan"1  y  =  tan'1  .f     ^  . 

1  +  zy 

If  the  numerical  values  of  the  functions  are  given,  as  0= 
and  tf  =  tan-1  J,  we  derive 


'  =  tan-1  %  +  tan"1  J  = 
and 


lrf-t 

55.  Examples. 

1.  Prove  the  following  identities: 

/   x  sin  :r-fsin  y  _ 
\a. 


/T  x  sin  a:  — sin  y 
(b)  -  — —  = 

y  cos  x  +  cosy 


2.  Simplify 


PLANE  TRIGONOMETRY.  49 

sin  3a  +  sin  5a 


COS  3a  — COS  5a 

c, .      T  - »    sin  x  +  sin  2x  +  sin  3x 

3.  Simplify-  — =-. 

cos  z-f-cos  2a;  +  cos  ox 

4.  Show  that  tan  (2  tan'1  a)  —  1        2  . 

x  —  a 

5.  If  tan'1  J  +  tan-1  J  +  tan'1  -J  +  tan-1  f  =  a,  find  a. 

6.  Solve  sin  2a;  =  2  cos  x. 

7.  Solve  cos  3x  —  cos  5z  =  sin  a:. 

8.  4  tan-1  J-tan'1  ^¥  =  ^  . 

9.  sin'1  f  +  sin-1  -fr  +  sin'1 1|  =  ^ . 

w 

12.  sin"1(cos  x)  H-cos"1(sin  y)  +x  +  y  =  7r. 

2 
x 


13.  tan-'  +tan"       _  +tan-     ,  =»,. 


50  PLANE  TRIGONOMETRY. 


CHAPTER  VI. 

FORMULA  FOR  THE  SOLUTION  or  OBLIQUE  TRIANGLES. 

56.  The  Sine  Formula. — Let  ABC  (Fig.  37)  be  any  oblique 
triangle,  and  from  the  vertex  C  drop  a  perpendicular  to  the  oppo- 
site side  c.  Let  A,  B  and  C  be  the  values  of  the  angles,  a,  ~b  and  c 
the  values  of  the  opposite  sides,  respectively,  and  p  the  value  of 
the  perpendicular. 

C 


From  the  two  right  triangles  BCD  and  ACD,  we  have 
p  =  asinB  =  b  sin  A. 

By  dropping  a  perpendicular  from  B  on  h,  we  obtain 
p'  =  a  sin  C=c  sin  A. 

From  these  two  equations, 


sin  A  "~  sin  B  ~~  sin  C  ' 
If  the  angle  A  is  obtuse,  the  perpendicular  falls  on  BA  pro- 

duced, and  then 

r> 

p  —  l  sin(180°—  A)  =b  sin  A  =  a  sin  2^° 

as  before.    Stated  in  words: 

The  sides  of  a  plane  triangle  are  proportional  to  the  sines  of 
their  opposite  angles. 


PLANE  TRIGONOMETRY. 


51 


57.  The  Cosine  Formula. — In  the  oblique  triangle  ABC  (Fig. 
38),  draw  the  perpendicular  p  from 
C  to  the  side  c;  then  in  the  right 
triangle  ACD  the  side  AD  =  b  cos  A, 
&JL,  in  BDC,  BD  =  a  cos  B=c-b 
cos  A.  Equating  the  value  of  p2 
from  each  of  the  two  right  triangles, 


from  which 


=  b2  +  c2-2bccosA. 


Similar  formulas  for  b2  and  c2  are  obtained  by  interchanging 
letters.    Hence 

The  square  of  any  side  of  a  triangle  is  equal  to  the  sum  of  the 
squares  of  the  other  two  sides  diminished  by  twice  the  rectangle 
of  these  sides  multiplied  by  the  cosine  of  the  included  angle. 

58.  The  Tangent  Formulae.—  From  Art.  56, 

a_  _  sin  A 

T  ~~  sin  B  ' 

and,  by  composition  and  division, 

a  —  b       sin  A  —  sin  B 


From  Art.  53, 


Hence 


a  +  b      sin  A  +  sin  B' 

sin  A  —  sin  B  _  tan^(A  —  B) 
sin  A  +  sin  B  ~  ta 


a-b  _tan|(A-B) 
a  +  b     .  tan|(A-f  B)  ' 


Similar  formulae  for  b  and  c,  and  a  and  c,  are  readily  derived  by 
interchanging  letters. 


52  PLANE  TRIGONOMETRY. 

59.  Functions  of  Half  -angles  in  Terms  of  the  Sides.  —  If  in  Art. 

57  we  put  cos  A  -  1  -  2  sin2  \A, 


.  2iA_a2-(b-c)2  _  (a-b  +  c)(a+b-c) 
~±fc~  ~^fc~ 

If  we  put  2s  =  (a+b  +  c),  then 

(a-b  +  c)=2(s-b),         (a+b-c)=2(s-c)  ; 
and,  substituting  these  values  in  the  above  equation,  we  obtain 


ac 


the  last  two  from  interchange  of  letters. 

60.  If  in  the  same  formula  we  put  cos  A  =  2  cos2  \A  —  1,  we  get 


or 

^2  -LA- 

COS      2-£i-  — 


Put  a+6  +  c=25;  then  (b  +  c  —  a)  =2(s  —  a),  whence 


ac 


The  last  two  are  written  by  interchange  of  letters. 


PLANE  TRIGONOMETRY.  53 

61.  Dividing  the  formulae  of  Art.  59  by  those  of  Art.  60  : 


s(s-c) 
62.  From  Art.  56,  we  may  obtain 

a  -f  b  _  sin  A  +  sin  B 

c  sin  C 

and 

a  —  b  _  sin  A  —  sin  B 

c  sin  C 

hence  by  49  and  53, 

a  +  b      sin 


c  sin  |(7  cos 

a-b      c 


c  sin  ^(  cos    ( 

But  A+B=180°-C,  and  J(A+£)  =90°--  |C;  and  thus 
sin  £  (A  +B)  =cos%C,  and  cos  %(A+B)  =sin|(7. 
Substituting  these  values,  we  obtain: 

a  +  b  _  cosl(A-B)  _  cos^(A-ff) 


c  sin^(7         ~  cos  %(A  +B)' 

a-b      sinl(A-B)  _  am$(A-B) 


c  cosJC         ~  sin 

63.  Area  of  an  Oblique  Triangle  in  Terms  of  Its  Three  Sides.  — 
We  have 

sin  A  =  2  sin  \A  cos  \A, 

and,  by  Arts.  59  and  60,  this  becomes 


sin  A  —  ^r-  ys(s-a)  (s  —  b)  (s—c). 
be 


54  PLANE  TRIGONOMETRY. 

Hence,  K,  the  area  of  the  triangle  (Fig.  37),  is 

K=^pc=^bc  sin  A, 
whence 

K=  Vs(s-a)  (s  —  l)  (s-c). 
We  also  have 

p=  —  \/s(s-a)(s-b)(s-c). 


PLANE  TRIGONOMETRY.  55 

CHAPTER  VII. 

SOLUTION  OP  PLANE  OBLIQUE  TRIANGLES. 
64.  Case  I:    Given  Two  Angles  and  One  Side,  A,  B  and  a. — 

The  third  angle  C  is  found  from  the  formula 

C=1SO°-(A+B). 
The  sides  are  found  by  the  sine  formula, 

I       sin  B       .    c       sin  C 

—  —  ~ — T  and    —  =  — — 7  j 
a       smA  a       smA 

or 

.     _  .     n  FIG.  39. 

o  =  a  sin  #  cosec  A,       c= a  sin  C  cosec  A. 

Example:    Given  A  =  50°  38' 50",  5=60°  7'  25",  a  =  412.67. 
(7  =  180° -(110°  46'  15")  =69°  13' 45". 

a=412.67  log  2.61560  log  2.61560 

A  =  50°  38'  50"  cosec  0.11168  cosec  0.11168 

J5  =  60°     r  25"  sin  9.93807-10 

(7  =  69°   13'  45"  sin  9.97082-10 


=  462.76  log  2.66535 


c  =  499.00  log  2.69810 

65.  Second  Method. — When  a  and  &  are  nearly  equal,  and 
greater  accuracy  is  desired,  we  may  write : 


a  sin  B  sin  A  —  sin  B 

:  —  r  =0,'  -  =  —  - 
sm  A  sin  A 


But 
whence 


, 
a  —  o  — 


sin  A 
or 


sin  A 


56 


PLANE  TRIGONOMETRY. 


Thus  in  the  preceding  example, 

i(5  +  A)=55°  23'     7.5" 
$(B-A)=  4°  44'  17.5" 
4  =  50°  38'  50" 


b-a  =   50.083 
a=412.67 
6  =  462.753 


cos  9.75439-10 
sin  8.91699-10 
cosec  0.11168 
log  2.91663 

log  1.69969 


66.  Case  II:    Given  Two  Sides  and  an  Angle  Opposite  One  of 
Them,  a,  b  and  A.  —  By  the  sine  formulae 


Having  found  B,  we  have 


and 

c=asin  C  cosec  A. 

The  angle  B  corresponding  to  the  com- 
puted sine  may  be  either  an  angle  in  the 
first  quadrant  or  its  supplement,  and  hence 
two  solutions  may  be  possible,  as  indicated 
in  the  geometrical  construction  of  Fig.  40. 
If  the  side  a  is  greater  than  b,  the  second 
point  of  intersection,  B2,  falls  on  the  side 
c  produced,  and  only  one  of  the  solutions 
fills  the  given  conditions,  as  the  triangle  AB2C  contains  180°  —  A. 
If  the  computed  value  of  sin  B  is  unity,  the  two  triangles  are 
two  coincident  right  triangles;  while  if  the  computed  value  of 
sin  B  is  greater  than  unity,  the  triangle  is  impossible. 

Example:    Given  A  =  27°  47'  45",  a  =  219.91,  6  =  251.33. 


PIG.  40. 


PLANE  TRIGONOMETRY.  57 

6  =         251.33       log      2.40024 

a  =         219.91       colog  7.65775—10     log     2.34225  log      2.34225 

A  =27°  47'  45"     sin     9.66869—10     cosec  0.33131  cosec  0.33131 

B,=  32°  12'  15"     sin     9.72668—10 

#2=1470  47' 45" 

Ol=l2Q°    0'    0"  sin     9.93753—10 

Cf=    4°  24'  30"  sin     8.88572—10 

c,=         408.40  log      2.61109  

C2=  36.248  log      1.55928 

Second  Method. — Having  found  B  from  its  sine,  we  may  com- 
pute AP=  b  cos  A  =  x,  and  PB^  =  a  cos  B  =  ±y,  the  negative  value 
resulting  from  the  value  of  B  in  the  second  quadrant,  whence 


67.  Case  III:     Given  Two  Sides  and  the  Included  Angle,  or 

a,  b  and  C.  —  From  the  tangent  formulae  (Art  58), 

B        a-b  . 


and  since!  (A  +B)  =4(180°  -C)  =  90°  -\C, 
tani(A+£)= 

Substituting  this  value,  we  derive 


«-6  co 


The  value  of  c  may  now  be  found  from  the  sine  formula,  or 
more  conveniently  from  the  formulae  of  Art.  62,  from  which 

-(a  +  M  c 
'c 
or 

c-(a-b)  s_ 
b)  s 

Example:    Given  a  =  6.2387,  &  =  2.3475,  (7=110°  32'. 


58  PLANE  TRIGONOMETRY. 

a—  b     —  3.3912  log      0.59008                                           log     0.59008 

a+6    —  8.5862  colog  9.06620— 10  log  0.93380 

l(A+B)=34°  44'     0"  tan     9.84092     10  cos  9.91477—10    gin    9.75569—10 

I  (A—  B)=17°  26'  33"  tan     9.49720     10  sec  0.02044           cosec 0.52324 

c  =  7.3962                                      log  0.86901 


c  —  7.3962  log     0.86901 

A  =52°  10'  33" 
B  =17°  IT  27" 

The  two  values  of  c  are  computed  here  for  a  check. 

68.  Second  Solution. — If  we  wish  to  find  c  only,  we  have,  by 
the  law  of  cosines  (Art.  57), 

c2  =  a2  +  b2-2abcosC, 

which,  however,  is  not  adapted  to  logarithmic  computation.    If 
we  put  cos  C  =  T-  2  sin2  %C, 


Let 


tan  0= 


a—  b 
and  the  radical  in  the  above  equation  becomes 


20=sec0, 
from  which 

c=  (a—  &)sec  6. 

Examples. 

1.  Given  a=. 062387,  6  =  . 023475,  0=110°  32',  find 
10'  33",  5  =  17°  ir  27",  c=. 073962. 

2.  Given  a=  35.237,  6  =  18.482,  0=110°  40'  30",  find 
49'  58",  B  =  22°  29'  32",  c  =  45.198. 

3.  Find  c  in  both  the  above  examples  by  Art.  68. 


PLANE  TRIGONOMETRY.  59 

69,  Case  IV:  Given  the  Three  Sides,  a,  b  and  c.  —  When  all 
three  angles  are  required,  the  formula  of  Art.  61  for  the  tangent 
of  the  half  -angle  is  the  most  convenient,  and  is  accurate  for  all 
values  of  the  angles.  When  only  one  angle  is  required,  the 
formulae  of  Arts.  4Sand  <WTiiay  be  used,  although  the  tangent 
formula  is  always  more  accurate  in  practice. 

The  formula  for  tan  \A.  may  be  written 

tan  p  =  -L. 

The  other  angles,  by  permuting  letters,  are  readily  obtained. 
Note  that  the  radical  expression  is  the  same  for  all  three  angles. 
Put  _ 

p_     /  (s-a)(s-b)(s-c)^ 

*  s 

and  the  formulae  become  : 

fcniA  =          ,    tan^=-,    tan  *(7=-. 


Example:    Given  a  =  6.8235,  6  =  5.2063,  c=  3.1628. 
a=   6.8235 
b=   5.2063 
c-   3.1628 


25=15.1926 


s=   7.5963  colog  9.11940-10 

8-a  =   0.7728  log  9.88807-10 

s-b  =  2.3900  log  0.37840 

s-c  =   4.4335  log  0.64675 

P2  log  0.03262 

P  log  0.01631 

iA  =  53°  20'  20"                    tan  0.12824 

lB  =  23°  28'  51"                    tan  9.63791-10 

10=13°   10'  49"                    tan  9.36956-10 

Check:     90°     0'  0" 


60 


PLANE  TRIGONOMETRY. 


D 

FIG.  41. 


70.  Second  Method.— When  a,  b 
and  c  are  given,  we  may  solve  as 
follows : 

Draw  the  perpendicular  p  from 
C  to  the  side  c,  and  call  the  seg- 
ments of  the  base  x  and  y,  as  in  Fig. 
41.  Then 


whence 
But 
and  thus 

whence 
and 


Examples. 

1.  Given  a  =  25.783,  6  =  37.686,  c  =  50.401;  solve,  by  Art.  59; 
4  =  29°  49'  22",  £  =  46°  37'  43",  (7-103°  32'  55". 

2.  Find  the  angles  in  Example  1  by  Article  70. 

71.  Solutions  of  Eight  Triangles  by  Using  Functions  of  Half- 
angles. — The  formulas  involving  sines  and  cosines  may  be  con- 
verted into  formulae  involving  functions  of  the  half -angles.  Thus 
in  the  case  of  the  right  triangle  (Art.  37)  when  the  hypothenuse 
and  one  leg  are  nearly  equal  and  accuracy  is  desired,  use  the 
following  method : 


PLANE  TRIGONOMETRY. 


ttl 


whence 


C-^=2  sin2 
c 

c  +  a  _0  _  o 


c  —  a 


c+a 


Since  b  =  Vc2  —  a?=  V(c— a)  (c+a),  these  formulae  furnish 
a  convenient  and  accurate  solution : 


Given 


a= 382.4 

c=383.2 

c-a=     0.8 


log  9.90309 
log  2.88400 


-10 


5  =  l°  51'     5" 
5=3°  42'  10" 
2 

6  =  24.748 


2)7.01909-10 
tan  8.50954-10 

log  2.78709 
log  1.39354 


72.  The  Radius  of  the  Circumscribed 
Circle. — The  center  of  the  circumscribed 
circle  (Fig.  4$)  lies  in  the  perpendicu- 
lars erected  at  the  middle  points  of  the 
sides.  From  the  figure  we  have 


sin  CO^sin  A  =        = 


whence 


FIG.  42. 


R  = 


2  sin  A ' 


62  PLANE  TRIGONOMETRY. 

From  Art.  63, 

2K 

smA  =  -=—. 
be 

Hence 

p_  ale 
~  4K' 

73.  Radius  of  the  Inscribed  Circle. — The  area  of  the  triangle 
ABC  (Fig.  43)  is  equal  to  the  area  of  the  three  triangles  having 


FIG.  43. 

the  sides  of  the  triangle  as  bases  and  the  radius  of  the  inscribed 
circle  as  altitudes.    Hence 


Examples. 

1.  Find  the  radii  of  the  inscribed  and  circumscribed  circles  of 
the  triangle  of  Art.  89. 

2.  Find  the  same  for  the  triangle  of  Example  1,  Art.  70. 

3.  Find  the  area  of  both  the  triangles  of  Examples  1  and  2. 


PLANE  TRIGONOMETRY.  63 


CHAPTER  VIII. 

COMPLEX  NUMBERS,  DE  MOIVRE'S  THEOREM,  DEVELOPMENT  OF 
SINE  AND  COSINE  IN  TERMS  OF  THE  ANGLE. 

74.  Complex  Numbers. — In  the  Algebra  the  subject  of  complex 
numbers  was  briefly  treated,  where  expressions,  like  a  +  bi  were 
subjected  to  certain  operations,  a  and  b  representing  real  num- 
bers, positive  or  negative,  i  the  so-called  imaginary  unit,  V  — 1. 
We  shall  here  show  how  such  numbers  may  be  graphically  repre- 
sented and  the  results  of  operations  performed  by  or  upon  them 
interpreted.    They  are  much  used  in  the  practical  computations  of 
rotating  machinery,  such  as  steam  turbines,  dynamos  and  motors. 

We  recall  first,  from  the  Algebra  (Arts.  3-5),  that  from  the 
definition  of  i—  V  —  1,  the  powers  of  i  are : 

i=if         *2=-l,         i3=-i,         t4=+l, 
i5=i,        ie=  -l,        i*=  -i,        i8=  +  1,  etc. 

Secondly,  that  if  we  have  an  equation  between  two  complex 
quantities  like 

a+bi=c-\-di, 

then  must  a  —  c  and  b  =  d;  in  other  words,  an  equation  containing 
real  and  imaginary  quantities  may  be  broken  up  into  two  equa- 
tions, one  containing  only  the  real  quantities,  the  other  only  the 
imaginary  quantities. 

75.  Graphical  Representation  of  Complex  Numbers. — If  we 
multiply  a  real  number  a  by  i,  and  this  product  again  by  i,  the 
result  is  —a,  or,  multiplying  a  twice  by  i  reverses  its  sign.  This  is 
graphically  shown  in  Fig.  44,  where  the  number  OP  =3  units  is 
changed  to  OP^—  —3  units,  by  a  change  of  180°  in  its  direction. 
In  other  words,  by  using  the  operator  i  twice  in  succession  as  a 


64 


PLANE  TRIGONOMETRY. 


multiplier,  the  line  OP  has  been  turned  through  two  right  angles. 
Multiplying  a  number  by  i  may  be  interpreted  as  turning  the 
direction  of  the  line  which  represents  it  through  one  right  angle, 


-X- 


-3 


25  <?|      3    7> 

.  -3-i 

Q. 
V 

FIG.  44. 

or  90° .  Thus  the  quantity  represented  by  the  line  OP,  becomes 
the  quantity  OQ,  when  multiplied  once  by  i;  and  for  this  reason 
we  take  the  axis  of  y  as  the  axis  of  imaginary  numbers,  just  as  we 
have  taken  the  #-axis  as  the  axis  of  real  numbers.  To  plot  a  com- 
plex number  like  3  +  4t,  for  instance  (Fig.  45),  lay  off  OQ  =  3 


PIG.  46. 


units  horizontally  on  the  axis  of  real  numbers,  and  the  4t  units, 
QP,  at  right  angles  (parallel  to  the  axis  of  imaginary  numbers, 
which  are  therefore  more  properly  called  quadrature  numbers). 


PLANE  TRIGONOMETRY.  65 

The  general  complex  number  x  +  yi  is  represented  in  Fig.  46, 
from  which  we  illustrate  the  following  terms  used  in  connection 
with  such  numbers.  The  vector  OP  —  r—Vx2  +  y2  is  called  the 
modulus  or  the  absolute  value  of  x  +  yi.  The  angle  XOP  —  6  is 
called  the  amplitude,  or  the  argument,  or  the  arc  of  x-\-yi. 

Modulus  of  (x+yi)  =r= 

Arc         of  (x  +  yi)  =  0=tsn\-1 

x 

76.  Polar  Coordinates  of  Complex  Numbers.  —  From  Fig.  46  we 
have 

x=rcos0,  y 


r2(cos2  0+sin2  6)  = 

and 

x  +  yi=r(cos  B+i  sin  0). 

This  trigonometric  expression  shows  more  clearly  the  vector  char- 
acter of  complex  numbers  and  leads  to  simpler  methods  in  their 
use. 

Examples. 

1.  Plot  the  complex  number  3  +  4i,  and  show  geometrically  and 
trigonometrically  that  the  angle  0  is  turned  through  90°  by  multi- 
plying the  number  by  i. 

Ans.     ^=tan-1(t),  ^/  =  tan-1(-f),  i.  e.,  0'  =  90°  +6. 

2.  Find  the  modulus  and  amplitude  of  the  following  complex 
numbers,  and  show  that  the  amplitude  of  each  is  increased  by  90° 
by  multiplying  by  i. 


3  +  3t,  -  ,        n       ,          -     .          - 

77.  De  Moivre's  Theorem.  —  The  theorem  of  De  Moivre  is  stated 
by  the  equation 

(cos  0+tsin  0)n  =  cos  n0  +  isin  n0, 
which  may  be  derived  in  the  following  way  : 


66  PLANE  TRIGONOMETRY. 

If  we  square  the  complex  number 

x  +  yi=z  =  r(cos  0+i  sin  0), 
we  get 


=  r2(cos2  0-sin2  0+2i  sin  0  cos  6) 
=  r2(cos20+tsin20). 

Multiplying  z2  by  2  =  r(cos0+tsin  6), 

zz  =  i*\  (cos  20  cos  0  —  sin  20  sin  0)  +  1  (sin  20  cos  0  +  cos  26  sin  0)  [• 
=  r3(cos30+isin30). 

If  this  process  be  continued  to  n  factors,  we  shall  get 
2n  =  rn(cos  0+i  sin  0)n  =  r"(cos  nQ+isinnO). 
Dropping  the  common  factorsrn, 

(cos  0+i  sin  0)  n  =  cos  n0+i  sin  nO. 

The  theorem  may  also  be  developed  by  multiplying  together  two 
expressions  like  (cosa+tsina)  and  (cos  /?  +  tsin/?)  and  noting 
that  the  product  is  <(  cos  (a  +/?)  +t  sin  (a  +(3)  [-,  and  so  on,  for  any 
number  n  of  angles;  finally  making  a=p=y,  etc.,  we  derive  the 
formula^  as  given. 

78.  Functions  of  Multiple  Angles.  —  If  we  write  De  Moivre's 
theorem, 

cos  n0  +  t  sin  nO=  (cos  0  +  t  sin  0)n, 

we  may  expand  the  second  member  by  the  binomial  theorem,  and 
equate  the  real  and  imaginary  terms  of  the  identity  (Art.  74). 
For  example, 

cos  20+i  sin  20=  (cos  0  +  i  sin  0)2 

=  cos2  0  —  sin2  0  +  i2  sin  0  cos  0, 
whence 

cos  20-  cos2  0-  sin2  0, 
isin20  =  i2  sin  0  cos  0, 

thus  establishing  the  formulae  of  Art.  8&r  U  (\ 


PLANE  TRIGONOMETRY.  67 

Ifw=3, 

cos  30  +  i  sin  30 

=  cos30  +  t3  cos2  0  sin  0  —  3  cos  0  sin2  0  —  tsin  3. 


Equating  real  and  imaginary  terms, 

cos  30=  cos3  0-3  cos  0  sin2  0, 
tsin30  =  i(3cos20sin0—  sin3  0), 

which  may  be  reduced  to 

cos  30  =4  cos3  0—  3  cos  0, 
sin  30=3  sin  0  —  4  sin3  0. 

Expanding  the  second  member  of  the  identity  and  equating  real 
and  imaginary  terms,  we  express  cos  nO  and  sin  nO  in  a  power- 
series  of  sine  0  and  cosine  0.  Noting  that  even  powers  of  i  are, 
real,  we  may  write  : 

cos  n0=  cosn  0-  2^=11  cos"-2  0  sin2  0 
« 

+  n(n-l)(n-2)(n-3)   cosn 
sin  n0=»  cos"-1  0  sin  0-  rc(n-l)(K,-2)  cosn-3  #  sin3  e 

L5 

+  tt(n-l)(tt-2)(n-3)(n-4)    cosn-5  e  sin5  #  +  § 

l£ 


Express  the  following  in  trigonometric  form  and  find  the  values 
of  the  first,  second,  third  and  fourth  powers  : 


7$.  Cosine  a  and  Sine  a  as  Power-series  of  a.  —  In  the  expres- 
sions for  cos  nO  and  sin  nO  of  the  preceding  article,  let  n0=a,  and 


68  PLANE  TRIGONOMETRY. 

let  n  increase  without  limit,  while  a  remains  unchanged.  Then 
0=  —  diminishes  without  limit;  and  at  the  limit  we  have  (Art. 
35). 

sin  0= sin—  =  —  ,  cos  0  =  cos—  =1. 

n]n=M         n'  njn=x 

Introducing  these  values  into  the  expansion  of  cos  nO  and  sin  nO 
(Art.  78),  they  become: 


n(n—l)      a2 

cosa=l f  *  — 7 

2  n2 


n(n-l)(n-2)(n-3) 

|4 


- 

n  [3  n3 

,    n(n-l)(n-2)(n-3)(n-4)  a5 
~[5~  "5F" 

When  n  increases  without  limit,  the  factors  containing  n  all 
vanish,  except  the  highest  power  of  n  in  each  term;  thus,  for 
example, 

n(n-l)(n-2)      ,       3       _21  , 

n3  "  n  +n2Jn=00 

The  expansions  then  become  : 


COSa  =    - 


a2  a4  a 


/T\ 


a  a  a  /TTx 

Sina-a-   -||-  +  -jr"-]j;  +  ..... 

Dividing  (II)  by  (I)   (synthetic  division), 


PLANE  TRIGONOMETRY.  69 

Examples: 


QO        -<         i    "     \          J-  I  W  \  -L 

cos  3   =  1  —   -        •  -T--  + 


a  =.052360. 

a2  =  .  002742^-  2  =  0.001371. 
as  =  -  000143^-  6  =  0.000024. 
24=  0. 


With  these  values,  we  obtain,  to  five  places  : 

sina=   .052360  -.000024  =  0.05234. 
cos  a  =  1  -  .001371  =  0.99863, 

tana=   .052360  +  .000048  =  0.05241. 

2.  Expand  ei0  by  the  exponential  theorem  (Alg.  Art.  161)  and 
by  putting  iO  for  x  in  the  expansion 


show  that 


=  cos#+isin  0. 
3.  Show  also  that  e~i9  =cos  0  —  i  sin  0,  and  hence  that 

cos  0=  — 

A 

and 


70  PLANE  TRIGONOMETRY. 

80.  De  Moivre's  Theorem  with  Fractional  Exponents. — If  we 

r\ 

replace  6  by  —  ,  in 
fi 

(cos  0+i  sin  6)n  =  cos  nO+isin  n0, 
we  get 

(cos — +  tsin — )    =  c 
V        n  n  / 


whence 


n  i\  j 

cos  —  +i  sin  —  =  (cos  0+i  sin  0)  n. 

IV  it 


Since  the  values  of  cos  6  and  sin  0  remain  unchanged  when  6  is 
increased  by  any  multiple  of  2?r? 

i 
(cos  0+i  sin  6)^  =-{cos(2 


We  may  therefore  derive  the  n  values  of  (cos  0+i  sin  6)  »T  by  mak- 
ing k  successively  equal  to  0,  1,  2,  3,  .  .  .  .  ,  (n  —  1)  . 

Examples. 

1.  Let  x+yi=8  =  8  (cos  0  +  t  sin  0)  =8  (cos  2for  +  isin  2&7r)  ; 
then   (x  +  yi)*  =  2  (cos  ~p  +  i  sin  ^)  .     Thence  show  that  the 

three  cube  roots  of  8  are  2,  —  l  +  tV%  and  —  1  —  tVs". 

2.  Find  the  three  roots  of  (  —  8)  and  plot  them;  verify  them  by 
multiplying  the  three  trigonometric  expressions  together. 


SPHERICAL  TRIGONOMETRY. 


71 


SPHERICAL  TRIGONOMETRY. 


CHAPTEE  IX. 

GENERAL  FORMULAE. 

81.  The  object  of  Spherical  Trigonometry  is  the  investigation 
of  the  relations  of  the  circular  functions  of  the  sides  and  angles 
of  a  spherical  triangle,  and  thus  finding  formulas  for  the  solution 
of  the  triangle. 

The  sides  and  angles  are  expressed  in  the  sexagesimal  units  of 
angular  measure,  and  their  numerical  values  are  independent  of 
the  size  of  the  sphere.  For  this  reason,  we  shall  consider  the 
radius  of  the  sphere  as  unity. 


PIG.  47. 


82.  Let  ABC  (Fig.  47)  be  a  spherical  triangle,  formed  on  the 
surface  of  the  sphere  by  the  intersection  of  three  planes  through 


72  SPHERICAL  TRIGONOMETRY. 

the  center  of  the  sphere,  0,  forming  the  triedral  angle  0  —  ABC. 
The  sides  of  the  triangle  a,  b  and  c,  measure  the  face  angles  of  the 
triedral  angle,  BO C,  AOC  and  AOB,  respectively,  while  the  angles 
between  the  three  planes  are  equal  to  the  angles  of  the  spherical 
triangle. 

A  plane  tangent  to  the  sphere  at  A,  and  thus  perpendicular  to 
OA,  intersects  the  three  planes  in  the  plane  triangle  ARC',  and 
the  angle  at  A  measures  the  angle  between  the  faces  AOB  and 
A 00,  and  thus  also  the  spherical  angle  BAC  —  A. 

In  the  right  triangle  OAB'  let  OA  —  \\  then  OB'  =  sec  c,  and 
AB'  =  i&n  c;  in  the  right  triangle  OAC',  0.4  =  1,  0C"  =  sec  &,  and 
AC'  =  ta,n  I;  let  x-ffC'. 

83.  Relation  between  the  Sides  and  an  Angle  of  a  Spherical 
Triangle. — We  have  from  plane  trigonometry,  in  the  triangles 
OVVvn&AVV: 

x2  =  sec2  b  +  sec2  c  —  2  sec  b  sec  c  cos  a, 
#2  =  tan2  &  +  tan2  c—2  tan  b  tan  ccos  A. 

Subtracting  the  second  equation  from  the  first,  and  noting  that 
sec2  B— tan2  0=1,  we  obtain 

0  =  2  +  2  tan  b  tan  c  cos  A  —  2  sec  b  sec  c  cos  a. 

Transposing  the  last  term,  and  multiplying  the  equation  by 
•J  cos  b  cos  c,  there  results 

cos  a=cos  b  cos  c  +  sin  b  sin  c  cos  A. 

Writing  similar  formulas  for  cos  b  and  cos  c,  we  have  the  funda- 
mental formulae  for  spherical  trigonometry: 

cos  a=cos  b  cos  c  +  sin  b  sin  c  cos  A,  "1 

cos  b  =  cos  a  cos  c  +  sin  a  sin  c  cos  B,  I  (A) 

cos  c  =  cos  a  cos  b  +  sin  a  sin  b  cos  C.  J 

84.  We  snail  assume,  in  accordance  with  the  principles  of  gen- 
erality involved  in  the  definition  of  the  circular  functions  of  plane 


SPHERICAL  TRIGONOMETRY. 


73 


trigonometry,  that  these  formulae  are  general  and  apply  to  spheri- 
cal triangles  whose  sides  or  angles  may  exceed  90° ; in  the  practical 
solution  of  spherical  triangles,  however,  we  shall  consider  those 
only  whose  parts  are  less  than  180°. 

85.  Relations  Existing  between  a  Side  and  the  Three  Angles 
of  a  Spherical  Triangle.— Let  A'B'C' 
be  the  polar  triangle  of  the  triangle  &' 

ABC  (Fig.  48). 

From    the    geometric    principles    of 
polar  triangles,  we  have : 


-6,     6'= 180°  -B, 

cr=i8o°-c,    c'=i8o°-a  * V 

From  Art.  83,  in  the  triangle  A'B'C', 

,  F  IG.   4o. 

we  have          *   tj       (L^Q  .  fa*  (L  -t   (* 

cos  a'  =  cos  V  cos  c'  +  sin  V  sin  c'  cos  A'. 

Replacing  the  values  of  a',  b',  c'  and  A'  by  the  values  given  above, 
^-cosA=  (—cosB)  (—cos  (7)  —sin  B  sin  C  cos  a. 

Changing  signs  of  all  the  terms,  and  interchanging  the  different 
letters,  we  get  the  three  formulae : 

cos  A  —  —  cos  B  cos  C+ sin  B  sin  C  cos  a^ 

cos  B=  —cos  A  cos  (7  + sin  A  sin  C  cos  6,  L  (B) 

cos  C  =  —  cos  A  cos  B  +  sin  A  sin  B  cos  c.  J 

86.  The  Sine  Formulae. — From  the  first  of  the  formulae  of  Art. 
83,  we  get 

A      cos  a  —  cos  b  cos  c 

cos  A  = ; — - — : —     — . 

sin  b  sin  c 

Squaring  both  sides,  and  noting  that 

1  —  cos2  A  =  sin2/!, 


74  SPHERICAL  TRIGONOMETRY. 

•  2  A  —  sin2  0  sin2  c  — cos2  a  — cos2  0  cos2  c  + 2  cos  a  cos  0  cos  c 

bin     A  —    ; r-^ ; — = —  . 

sin  *  0  sin  c 

Putting  for  sin2  b  sin  2  c  its  equal,  (1  — cos2  0)  (1  — cos2  c),  we  get 
sin2  b  sin2  c=l  —  cos2  0  — cos2  c  +  cos2  b  cos2  c, 

whence 

.  2  ,      1  — cos2  a  — cos2  0  — cos2  c+2  cosacos  0  cos  c 
sin2  A  =  -  — .          .    —  -  . 

sin2  0  sin2  c 

Dividing  both  sides  by  sin2  a, 

sin2^.  _  1  —  cos2  a  —  cos2  b  —  cos2  c  +  2  cos  a  cos  &  cos  c  _  -, 
sin2  a  ~  sin2  a  sin2  b  sin2  c 

a  constant,  since  the  right-hand  side  of  the  equation  remains  un- 
changed, when  A  and  a  are  replaced  by  either  B  and  b,  or  C  and  c. 
Hence, 

sin2  A  _  sin2  B  _  sin2  C 

sin2  a  •  sin2  b  ~  sin2  c  ' 
or 

sin  ^4.        sin  B    _  sin  (7 

sin  a  " "  sin  0  " "  sin  c  ' 
or 

sin  a  sin  B= sin  &  sin  A,! 

sin  a  sin  (7 = sin  c  sin  A,  i-  ( C) 

sin  b  sin  C  =  sin  c  sin  B.  J 

Stated  in  words,  these  formulae  read : 

In  any  spherical  triangle,  the  sines  of  the  angles  are  propor- 
tional to  the  sines  of  the  opposite  sides. 

87.  The  sine  formula  may  be  derived  from  a  geometrical  figure 
(Fig.  49) .  From  any  point  P  in  OC,  draw  PM  perpendicular  to 
the  plane  OAB,  and  draw  MA'  and  MB'  perpendicular  to  OA  and 
OB,  respectively. 


SPHERICAL  TRIGONOMETRY. 


75 


Then  the  angle  PA'M  =  the  angle  A,  and  PB'M=ihe  angle  B. 
PM  PM 


hence 


BinA  =  Pl" 

sin  A  _  PBf 
sin  B  ~  PA' ' 


(1) 


FIG.  49. 


In  the  triangles  POA'  and  FOR,  the  angle  POA'  =  lt  and 
POB'  =  a;  and 


sma= 


PA' 


and 


FO  ' 
sin  a       PB' 


sin  b  ~  PA' 
From  (1)  and  (2), 


or 


sin  A  _  sin  a 
sin  J5       sin  & 

sin  A  _  sin  B 
sin  a       sin  ft 


76 


SPHERICAL  TRIGONOMETRY. 


88.  Formulae  for  the  Half  -angles.  —  In  the  first  of  the  formulae 
(A),  put 

cos  A  =  1  —  2  sin2  \A.  ; 
then 

cos  a—  cos  b  cos  c  +  sin  b  sin  c  —  2  sin  b  sin  c  sin2  \A,        (I) 

whence,  since 

cos  b  cos  c  +  sin  b  sin  c=cos(&  —  c), 

(2) 


.. 
2  sm  b  sin  c 

From  Art.  53,  Plane  Trigonometry, 
cos  (6  —  c)  —  cosa=  —  2  sin  i(6  —  c  +  a)sin£(&  —  c—  a) 


Let 
then 

and 

J(a+c  —  &)  =s  —  b. 

Substituting  the  values  of  (3)  and  (4)  in 


Similarly, 
and 


sin2  1?  A  _sin  (s-fr)  sin  (s-c) 
sin  &  sin  c 


sin2  ig-sin  (s-ft)  sin  (s-c) 
sin  a  sin  c 


.n2 


sin  a  sm 
89.  Substituting  in  the  same  formula  of  Art.  83, 

cos  A  =  2  cos2  \A.  —  1, 

cos  a  —  cos  b  cos  c  —  sin  b  sin  c  +  2  sin  b  sin  c  cos2  \A. 
=  cos  (  &  -f  c)  +  2  sin  b  sin  c  cos2  JA, 


(3) 


(4) 


SPHERICAL  TRIGONOMETRY. 


77 


whence 


C^S  ifxl  — : ; : ~    y 

2  sin  b  sin  c 

cos  a  —  cos  ( b  +  c)  =  —  2  sin 
=     2  sin 

Introducing,  as  in  Art.  88,  s  = 
Similarly, 


a—  I  —  c) 
— a). 


sin  6  sm  c 


sn  a  sin  c 


(E) 


sin  a  sin  b 
90.  The  quotient  of  (D)  divided  by  (E)  gives: 


tana*A  = 


sin  (5  —  &)  sin  (s  —  c} 
sin  s  sin  (s  —  a) 


sin  s  sin  (  s  —  b  ) 


tan 


sin  5  sin  (s  —  c) 


91.  Formulae  for  the  Half  -sides.  —  To  find  similar  formulae  for 
the  half  -sides,  apply  the  formulae  (D),  (E)  and  (F)  to  the  polar 
triangle,  thus  : 


=  90°— 


c=180°-C" 


=  270°-^ 


Whence  we  have 


s-a= 
5-6= 


sn        =  cos 


cos 


=  sin2  Ja',     tan2  $A  =  cot2  Ja'. 


78 


SPHERICAL  TRIGONOMETRY. 


Substituting  the  above  values  in  the  formulae  (D),  (E)  and  (F), 
and  dropping  primes,  we  obtain 


sm  B  sin  G 


tan2  la  • 
1     fl- 


sin  B  sin  C 
-™* 


92.  Formulae  of  Gauss  or  De  Lambre,  —  From  (D)  and  (E), 


sn       cos       = 


sm  c 


/sm 
Y      si 


sin  a  sin  6 


or 


sm  c 


cos4(7. 


Similarly, 

cos  \A.  sin  45  = 

cos  if  A  cos  45  = 
sin   4  sin    5= 


sm  c 


sm  c 


cos 


sn 


(1) 

(2) 
(  3  ) 
(4) 


From  Plane  Trigonometry  (Art.  4^),  (1)  +  (2) 


gives 


sine 


Reducing, 
Similarly, 


cos  4c  sin 


sm  \c  cos  %c 
^  cos      a  — 


(5) 


sm  c 


2  sm  -|c  cos  ^c 
sin  Jc  sin  4  (A—  5)  =0034^  sin  4  (a—  &). 


(6) 


SPHERICAL  TRIGONOMETRY.  79 

In  the  same  way, 

1  /  A  ,  T>\       •    in  f  -sins  —  sin  (s—  c)  \ 
cos4(A+£)=sm4(7|  --  ^  --  ) 

=  |^l(^+&)smiC| 

I        2  sin  Jc  cos  Jc     J 

or 


and,  similarly, 

sinico>si(A--#)=siiil<7sin4(a+&).  (8) 

Collecting  (5),  (7),  (6),  and  (8),  we  have  Gauss'  Equations: 


cos    c  sn 

cos  Jc  cos  ^  (A  +  B)  =8in4(7co8j(o-f-J), 

sin  Jc  sin  J  (A—  5)  =cos-J(7  sinj(a—  6),  | 
sin  \c  cos  ^  (A—  5)  ^sinJC'  sini(a+&).j 

93.  Napier's  Analogies   (or  Eatios)  .—Dividing   (6)   by   (5) 
(Art.  92),  and  transposing, 


tan^c 
From  (8)  -(7), 


_  cos  ^  (A  -B) 
tan|c  c 

From  (6)  -7-  (8), 


From  (5)  -(7), 

tan  \  (A  +B]  _ 
cot  4(7        ~ 


80  SPHERICAL  TRIGONOMETRY. 


CHAPTER  X. 
SOLUTION  OF  SPHERICAL  RIGHT  TRIANGLES. 

94.  When  one  of  the  angles  of  a  spherical  triangle  is  a  right 
angle,  the  general  formulas  of  Chapter  IX  assume  forms  which 
may  be  generalized  under  two  simple  rules. 

Assume  the  angle  0  =  90°,  and  the  formulae  of  (A),  (B)  and 
(C)  become: 

FromA(3),  cos  c=cosacos  b.-  (1) 

"     B(3),  cosc=cot^cot#.  (2) 

"     B(l),  cos  A  =  cos  a  sin  B.  (3) 

"     B(2),  cos  B= coat  sin  A.-  (4) 

"     C(2),  sin  a = sin  c  sin  A.  (5) 

"     0(3),  sin  6  =  sine  sin  B.  (6) 

From  the  triangle  AB'C'  (Fig.  47), 

cos  A  =  ^^  =  tan  b  cot  c,  (7) 

tan  c 

cos  B=  tan  a  cot  c,  ( 8 ) 

by  interchanging  a  and  b  and  A  and  B  in  (7). 

We  have  tan  A  =sm      ,  and  substituting  the  values  of  sin  A 
cos  A 

and  cos  A  from  (5)  and  (7), 
tan  A  — 


tan  b  sin  c       cos  c  tan  b  3 
and  putting  cos  c=  cos  a  cos  I,  from  (1),  we  obtain,  finally, 


cos  a  sm         sin 
or 

sin  b  =  tan  a  cot  A  ;  (9) 

and  by  interchange  of  letters, 

sin  a  =  tan  b  cot  B.  (10) 


SPHERICAL  TRIGONOMETRY.  81 

These  ten  formulae  are  sufficient  for  the  solution  of  any  spherical 
right  triangle  of  which  any  two  parts  are  given  (excluding  the 
right  angle). 

95.  Napier's  Rules. — The  ten  formula?  necessary  for  the  solu- 
tion of  any  right  triangle  from  two  given  parts,  may  be  readily 
derived  from  two  artificial  rules,  devised  by  Lord  Napier. 

Omitting  the  right  angle,  and  taking  the  complements  of  the 
hypothenuse  and  of  the  two  adjacent  angles,  the  five  parts  are 
called  the  circular  parts  of  the  right  tri- 
angle, and  are  lettered  as  in  Fig.  50. 

If  we  consider  any  three  of  these  circular 
parts,  one  of  them  is  the  middle  part  and 
the  other  two  either  adjacent  or  opposite 
parts.  For  example,  of  the  parts  b,  a  and  C0i 
co-B,  a  is  the  middle  part  and  b  and  co-B  \ 
adjacent  parts,  the  right  angle  being 
omitted;  while  of  the  three  parts  co-A,  co-B  and  b,  co-B  is  the 
middle  part  and  b  and  co-A  opposite  parts,  since  each  is  separated 
from  co-B  by  an  intervening  part. 

Napier's  two  rules  are : 

1.  The  sine  of  the  middle  part  is  equal  to  the  product  of  the 
tangents  of  the  adjacent  parts. 

2.  The  sine  of  the  middle  part  is  equal  to  the  product  of  the 
cosines  of  the  opposite  parts. 

Example:  Given  a  and  b,  to  find  the  formulae  for  the  other 
parts  of  the  triangle  in  terms  of  functions  of  the  given  parts  alone. 

To  find  the  angle  A :  The  three  circular  parts  are  a,  b  and  co-A; 
b  is  the  middle  part;  and  the  other  two  adjacent  parts;  hence 
sin  b  =  tan  a  cot  A,  or 

cot  A  =  sin  b  cot  a.  (1) 

To  find  B:  a  is  the  middle  part  and  b  and  co-B  adjacent  parts ; 
hence  sin  a  —  tan  b  cot  B,  or 

cot  B= sin  a  cot  b.  (2) 


82  SPHERICAL  TRIGONOMETRY. 

To  find  c:  co-c  is  the  middle  part,  a  and  b  opposite  parts; 
hence 

cos  c  —  cos  a  cos  6.  (3) 

A  check  formula  involving  the  three  required  parts,  co-c,  co-A 
and  co-B,  is  found  by  taking  co-c  as  the  middle  part,  co-A  and 
co-B  as  adjacent  parts,  giving 

cos  c— cot  A  cot  B.  (4) 

Note  that  the  check  formula  is  always  expressed  in  terms  of  the 
functions  l>y  which  the  parts  are  obtained. 

96.  Solution  of  Spherical  Right  Triangles. — When  two  parts 
of  a  spherical  right  triangle  are  given  (not  including  the  right 
angle),  each  of  the  other  parts  is  to  be  determined  in  terms  of  the 
functions  of  the  two  given  parts  only,  by  means  of  Napier's  rules, 
and  the  computation  checked  by  a  "  check  formula  "  involving 
the  three  computed  parts. 

It  is  customary  to  divide  the  solution  into  five  cases,  according 
to  the  parts  given;  but  this  is  not  necessary,  as  Napier's  rules 
furnish  an  obvious  method  for  any  case  that  may  arise. 

97.  To  Determine  the  Quadrant  in  Which  a  Computed  Part 
Belongs. — If  the  computed  part  is  found  from  its  cosine,  tangent 
or  cotangent,  the  sign  of  the  function  determines  the  quadrant; 
but  if  it  is  found  from  its  sine,  it  may  be  in  either  the  first  or 
second  quadrant.    Except  in  the  particular  case  of  a  double  solu- 
tion, the  proper  quadrant  for  a  part  found  from  its  sine  may  be 
determined  by  one  of  the  two  following  principles. 

98.  1.  In  a  right  spherical  triangle  an  angle  and  its  opposite 
side  are  always  in  the  same  quadrant. 

For  we  have 

cos  A  =  sin  B  cos  a; 

and  since  sin  B  is  always  positive  (#<180°),  cos  A  and  cos  a 
must  have  the  same  sign;  that  is,  A  and  a  must  be  either  both 
less  or  both  greater  than  90°. 


SPHERICAL  TRIGONOMETRY.  83 

99.  2.  When  the  two  sides  including  the  right  angle  are  both 
in  the  same  quadrant,  the  hypothenuse  is  less  than  90° ;  when  the 
two  sides  are  in  different  quadrants,,  the  hypothenuse  is  greater 
than  90°. 

This  follows  from  the  relations  between  the  three  sides, 

cos  c^cos  a  cos  b. 

For  if  a  and  b  are  in  the  same  quadrant,  cos  a  and  cos  b  have  like 
signs  and  cos-c  is  positive,  so  that  c  is  in  the  first  quadrant;  i.  e., 
less  than  90°.  If  a  and  b  are  in  different  quadrants,  their  cosines 
have  different  signs ;  cos  c  is  therefore  negative,  and  c  is  in  the 
second  quadrant ;  i.  e., £>9^j^ 

100.  Double  Solutions. — When  the  two  given  parts  are  a  side 
and  its  opposite  angle,  there  are  always  two  solutions,  as  in  such  a 
case  the  required  parts  are  all  found  from  their  sines,  and  may  be 
therefore  in  either  the  first  or  second  quadrant.     In  arranging 
and  marking  the  parts  which  go  together,  attention  must  be  paid 
to  the  principle  of  Art.  99. 


101.  Examples. 

Example  1:  Given  a  =  37°  45'  20", 
b  =  113°  10'  30".  Taking  the  two  parts 
a  and  b  with  each  of  the  other  three  in 
turn  as  the  third  part,  we  find,  Fig.  51 :  Fio.^51. 

cos  c= cos  a  cos  b;  (1) 

sin  b  =  tan  a  cot  A,  or  cot  A  =  sin  b  cot  a;  (2) 

sin  a = tan  b  cot  5,  or  cot  B— sin  a  cot  b;  (3) 

Check:  cos  c  =  cot  .4  cot  #.  (4) 

Note  that  the  check  formula  for  a  right  triangle  always'  con- 
tains the  functions  from  which  the  required  parts  should  be  found. 


84  SPHERICAL  TRIGONOMETRY. 


=  37  45  20     cot   0.11101  sin     9.78696  cos     9.89798 

=113  10  30     sin    9.96340  cot     9.63153n  cos     9.59499w 


A—  40     6  42     cot   0.07447 


41  15     .     - cot     9.41849n 


c  =108     7  43 cos    9.49297?! 

Check:   (cot  A  cot  B)=cos     9.49296n 

Example  2:    Given  &  =  154°  7',  #=152°  23'. 
Formulae :   ( 1 )   sin  a = tan  b  cot  5, 

(2)  sin  A  =  sec  &  cos  Bf 

(3)  sin  c=sin  b  cosec  5, 

(4)  sin  a= sine  sin  A  (check). 

o          /        " 

6   =154    7     0    tan     9.68593ft  sec      0.04591ft          sin      9.64002 

B  =152  23     0     cot    0.28127w  cos      9.94747ft          cosec  0.33390 


ax  =  68  02  36     sin     9.96730 

a2=lll  57  24  

A,=  80     1  20 sin       9.99338 

Ao=  99  58  40 


c2  =  70  20  30  sin  A  9.99338 

Check:  sin  a  9.96730 

For  the  proper  grouping  of  the  parts  (Kules  1  and  2),  a±  and  A^ 
are  put  in  the  same  quadrant,  the  first;  and  c^  must  be  in  the 
second  quadrant,  since  #±  and  b  are  in  different  quadrants. 

Derive  the  formulae  and  check,  and  solve  the  following  right 
triangles : 

o          /        "  o          '       "  o          '          " 

3.     c=109  41  50  4.     o=  67     6  53  5.     A=  55  18  13 

6  =  25  52  34  A=  80  10  30  c  =  75  20  30 

6.     A=138  27  18  7.     a  =116  24  25  8.     6=154     032 

B—  80  55  27  6=  16  50  30  5=126  57  37 

9.     6=60  10.     a=  90  11.    a=  90 

A=  90  6  =  60  6  =  90 


SPHERICAL  TRIGONOMETRY.  85 

102.  Additional  Formulae  for  the  Solution  of  Spherical  Right 
Triangles.  —  In  cases  where  a  part  is  found  from  its  computed  sine 
or  cosine,  the  formulae  of  the  preceding  articles  may  have  to  be 
transformed  into  formulae  involving  functions  of  the  half-angle 
to  insure  sufficiently  accurate  results. 

Thus,  if  we  have  given  c  and  a,  we  have 

.    A  _  sin  a 

8111  /t  —  —  -  , 

sin  c 
from  which 

1  —  sin  A  __  sin  c  —  sin  a 
1  +  sin  A  ~  sin  c  +  sin  a  ' 


sin  c  —  sin  a=2  cos  -J(c+a)sin  -J(c—  a), 
sin  c  +  sina=2  sin  J('c+a)cos£(c—  a). 

Hence 

tnn«(45"-iA)= 


From  cos  c=cos  a  cos  &,  or  cos  &=  cos  c  ,  we  derive 

cos  a 

1  —  cos  &  _  cos  a—  cos  c 
1  +  cos  6  ~~  cos  a  +  cos  c  ' 

or 


From  cos  B= 


tan  c ' 

1  —  cos  B  _  tan  c  —  tan  a 
1  +  cos  L  ~  tan  c  +  tan  a  ' 

tan2iB=^r/"  ^ 


86  SPHERICAL  TRIGONOMETRY. 

Examples. 
1.  Derive,  from  cos  c—  cot  A  cot  B,  the  formula 


2.  From  cosa=?3  derive 


103.  Quadrantal  Triangles.  —  If  one  side  of  a  spherical  triangle 
is  90°,  it  is  called  a  quadrantal  triangle.  In  its  polar  triangle,  the 
angle  opposite  the  quadrantal  side  is  90°  ;  and  the  formulae  for  the 
right  triangle  thus  formed  are  easily  transformed  into  those  for 
the  given  triangle.  Generally  but  one  or  two  parts  of  the  triangle 
are  required  and  formulae  (A),  (B),  (C)  are  readily  converted 
by  putting  the  given  side  of  90°  in  the  proper  formulae. 


SPHERICAL  TRIGONOMETRY. 


87 


CHAPTER  XI. 
SOLUTION  OF  OBLIQUE  SPHERICAL  TRIANGLES. 

104.  Any  oblique  spherical  triangle  may  be  divided  into  two 
right  triangles  by  a  perpendicular  from  an  angle  to  its  opposite 
side,  and  the  complete  solution  of  the  triangle  obtained  in  terms 
of  the  given  parts  and  certain  parts  of  the  right  triangles  as 
auxiliaries,  thus  reducing  the  solution  of  all  spherical  triangles 
to  the  application  of  Napier's  rules  for  right  triangles. 

When  the  three  given  parts  are  the  three  sides  (a,  b,  c)  or  the 
three  angles  (A,  E,  C],  however,  the 
solution  by  the  formulae  for  the  half- 
angle  is  much  more  expeditious  and 
accurate. 

In  solving  other  cases,  draw  a  plane 
triangle  to  represent  the  spherical  tri- 
angle, and  letter  the  segments  of  the 
base  <f>  and  <£',  and  the  parts  of  the 
opposite  angle  6  and  #',  as  in  Fig.  52.  The  perpendicular  p  is 
eliminated  in  the  solution  and  for  that  reason  is  not  lettered.  The 
other  parts  of  the  triangle  are  to  be  so  lettered  that  (1)  two  of  the 
given  parts  shall  be  in  the  first  triangle  and  (2),  whenever  pos- 
sible, two  of  the  required  parts  in  the  second  triangle. 

It  will  be  found,  that  when  Ajbp^OT  AJ3f  ,(two  parts  including 
a  third  part)  are  given,  there  is  a  choice  of  two  ways  in  which  the 
diagram  may  be  lettered;  and  that  one  should  be  taken  which 
brings  the  two  required  parts,  when  only  two  are  required,  in  the 
second  triangle. 


PIG.  52. 


88 


SPHERICAL  TRIGONOMETRY. 


When  two  parts  and  a  part  opposite  one  of  them  are  given,  the 
perpendicular  can  be  drawn  in  but  one  way. 

There  is  no  ambiguity  about  the  quadrant  in  which  the  com- 
puted parts  belong,  as  the  two  rules  concerning  right  triangles 
(Arts.  98  and  99)  apply  to  both  the  right  triangles. 

Note  that  in  double  solutions  0  is  in  the  same  quadrant  as  $,  & 
in  the  same  quadrant  as  <f>'. 

105.  Case  I:    Given  A,  b,  c,  required  E,  a — Let  two  of  the 

parts  of  the  first  triangle  be  A 
and  b,  as  in  Fig.  53,  making  the 
two  required  parts  fall  undivided 
in  the  second  triangle. 

From   the   first   triangle,   by 
Napier's  rules, 

tan  <f>  =  cos  A  tan  ~b,          (1) 
*'=c-$.  (2) 

To  find  a:    Associate  a  and  <j>'  with  b  and  <f> : 

cosa=cos<f>'  cosp, 
cos  &  =  cos  <f>  cos  p. 


Eliminating  p  by  division, 

cos  a = cos  b  cos  </>'  sec  </>. 

To  find  B:    Associate  B  and  </>'  with  A  and  <f> : 

sin  <£'  =  cot  B  tan  /?/ 
sin  4>  =  cot  A  tan  p. 


(3) 


Eliminate  tanp: 


cot  B  =  cot  A  sin  <f>'  cosec  <£, 
cos  B  =  cot  a  tan  <f>'  (check). 


(4) 
(5) 


SPHERICAL  TRIGONOMETRY.  89 

If  the  angle  C  is  also  required,  we  find  6  and  tf,  using  the  given 
parts  and  the  two  auxiliary  parts  <f>  and  <f>'. 

cot  0= cos  &  tan  A.  (6) 

sin  p  =  cot  &  tan  <£', 
sin  p  =  cot  0tan<£. 

Eliminating  sinp, 

cot  tf^cot  6  tan  <£  cot  <£',  (7) 

0=0+0*.  (8) 

The  whole  computation  may  be  checked  by  the  sine  formula, 

sin  A  _  sin  B  _  sin  C  /o^ 

sin  a        sin  6        sin  c  * 

106.  If  tan  (/>  from  (1)  is  negative,  and  A  is  obtuse,  the  per- 
pendicular will  fall  to  the  left  of  A  on  the  side  BA  produced,  in 
which  case  <£  will  be  negative  and  less  than  90°  ;  this  case  is  shown 
in  Pig.  54. 


M    fl    A    C     0  tf  fa 

FIG.  54.  FIG.  55. 

If  tan  (f>  from  ( 1 )  is  negative,  and  A  is  acute,  the  perpendicular 
will  fall  to  the  right  of  A  (either  upon  AB  or  AB  produced),  in 
which  case  <f>  will  be  positive  and  lie  in  the  second  quadrant ;  this 
case  is  illustrated  in  Fig.  55. 

The  same  result  will  be  reached,  however,  if,  in  all  cases,  the 
auxiliary  <f>  is  taken  in  the  first  or  second  quadrant  according  to 


90  SPHERICAL  TRIGONOMETRY. 

the  sign  of  tan  </>,  and  the  proper  signs  of  the  functions  of  the 
auxiliary  arcs  <f>  and  c  —  (/>  are  observed. 

107.  Case  I :  Second  Solution,  Given  A,  6  and  c. — This  prob- 
lem is  of  importance  in  navigation,  where,  however,  only  the  side 
a  and  the  angle  B  are  required. 

The  angle  B  is  then  usually  taken  from  "  azimuth  tables,"  com- 
puted for  that  purpose,  and  the  side  a  is  computed  from  the 
formula 

cos  a = cos  &  cos  c  +  sin  &  sin  c  cos  A, 

after  substituting  1  —  2  sin2  ^A  for  cos  A,  giving 

cosa=cos(&  —  c)  —2  sin  &  sin  csin2|A;  (10) 

or  else  by  putting  cos  A  =  1  —  vers  A.,  giving 

cosa=cos(&  —  c)  —sin  6  sin  c  vers  A.  (11) 

The  following  example  is  solved  by  both  methods,  as  well  as  by 
the  formulae  of  Art.  93.  cfc' 

Note,  in  the  first  solution,  tan  <f>  is  negative  an<Lis  taken  in  the 
second  quadrant,  where  in  reality  it  belongs  from  the  given  parts. 
If,  however,  <f>  is  taken  as  a  negative  angle,  its  value  would  be 
-  (59°  28'  39") ;  <£'  becomes  124°  57'  9".  Following  these  parts 
through  the  computation,  we  get  A  and  B  from  their  functions, 
in  agreement  with  the  results  already  obtained  by  taking  </>  in  the 
second  quadrant.  Cot  9  will  be  negative  as  before,  and  will  have 
to  be  taken  as  a  negative  angle,  to  be  in  the  same  quadrant  as  <f> ; 
cot  &  will  still  be  negative  and  must  be  taken  in  the  second  quad- 
rant with  (j>f ;  thus 

0  =-(68°  49'  23") 
&-     114°  40'     9" 

e+tf=       45°  50'  46" 
as  before. 


SPHERICAL  TRIGONOMETRY. 


91 


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SPHERICAL  TRIGONOMETRY. 


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SPHERICAL  TRIGONOMETRY. 


93 


1.  Given  A  =  50' 
6  =  59 
c=109 


2.  Given  5=102' 
a=  98 
c=  99 


Examples. 

10'  10" 
29    30 

39    40     find  a,  B,  and  C. 
Ans. 


55'     4" 

8  18 

9  48     find  &,  A,  and  (7. 

Ans. 


a=   69°  34'  56" 
#=   44     54    31 
(7  =  129     29    52 


find  c,  A,  and  B. 
Ans. 


17'  58" 

16    48 

6    20 

3.  Given  C  =  99°  44'  46" 
a=81       2    30 
6  =  99     58    38 

3'     0" 
43      8 
30    24 

111.  Case  II:  Given  A,  B  and  c,  to  Find  C  and  a. — Letter  the 
diagram  (Fig.  56)  so  as  to  bring  the  required  parts  (7  and  a  in  the 
second  triangle.  Here  one  of  the  given  angles,  B,  is  divided,  and 
we  may  first  deduce  6  and  ff. 


A  =  100 
(7=101 


c=101( 

A=   82 
B=   98 


cot  0  =  cos  c  tan  A, 

^  =  £-0, 

cot  a=cot  c  cos  0'  sec  0, 
cos  (7  =  cos  A  sin  0'  cosec  0, 
cos  a = cot  6'  cot  (7  (check). 


(3) 
(3) 
(4) 
(5) 


94  SPHERICAL  TRIGONOMETRY. 

If  b  is  required, 

tan  <£  =  cos  A  tan  c,  (6) 

tan  <£'  =  tan  <£  cot  0  tan  0',  ( 7 ) 

&:=<£  +  </>'.  (8) 

The  whole  computation  is  to  be  checked  by  the  sine  formula. 
The  problem,  however,  never  comes  in  navigation,  and  very  sel- 
dom, if  ever,  in  any  practical  way. 

Examples. 

1.  Given  1  =  115°   56'  23" 

B=   50     51    43 
c=   52     29    24     find  a,  b,  and  C. 

Ans.     a=89°  29'  30" 
6  =  59     35    50 
£$'  (7  =  45     30    35 

2.  Given  b  =   93°  -96'     2" 

A  =  97     17    38 

0  =  100     10    54     find  5,  a,  and  c. 

Ans.     5  =  94°  48'  16" 
a  =  96     33    58 
c  =  99     40    34 

3.  Given  a=   99°  40'  48" 

5=114     26    50 

C=   82     33    31     find  A,  b,  and  c. 

Ans.     A=   95°  38'     4" 
6  =  115     36    45 
c=   79     10    30 

112.  Case  III:  Given  a,  b  and  A,  to  Find  B  and  c. — The  per- 
pendicular can  be  drawn  in  only  one  way,  as  in  Fig.  57. 

tan  <£  =  cos  A  tan  b.  (1) 

cos  a=  cos  <£'  cos  p,~] 
cos  &  =  cos  <£  cos  p.  J 


SPHERICAL  TRIGONOMETRY.  95 

Eliminating  cos  p, 

cos  </>'  =  cos  (j>  cos  a  sec  &,  (2  ) 

c  =  *±*',  (3) 

sin  <£'=cot  B  tan  p,' 

sin  </>  —cot  A  tan£>. 

Eliminating  tan  p, 

cot  J5—  cot  A  sin  <£'  cosec  </>.  (4) 

It  is  important  to  note  that  <j>',  being  found  from  its  cosine,  may 
be  either  a  positive  or  a  negative  arc,  and  hence  there  are  two  solu- 
tions possible.  This  is  always  the  case  in  spherical  triangles 
where  two  of  the  given  parts  are  an  angle  and  a  side  opposite. 

This  problem  is  of  frequent  occurrence  in  navigation,  where, 
however,  only  the  parts  B  and  c  are  required. 

If  the  angle  C  were  required,  it  would  be  computed  from  the 
formulae 

cot  0  =  cos  b  tan^l,  (6) 

cot  ef  =  cot  0  tan  <f>  cot  <£',  (7) 

(8) 


Check  by  the  sine  formula. 

113.  The  following  example  of  Case  III  is  solved  completely 
and  checked  : 


96 


SPHERICAL  TRIGONOMETRY. 


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SPHERICAL  TRIGONOMETRY. 


97 


I.  Given  a=99c 

6  =  64 

4=95 


40'  48" 
23    15 

38      4 


Examples. 


find  E,  c,  and  (7. 
Ans. 


B=   65° 
c=100 


33' 

49 


9" 

28 


C=   97     26    26 


2.  Given  a= 

6  = 

A= 


40 


29 


find  B,  c,  and  C. 
Ans. 


37'  20" 
38    42 

27 
40 
40 


1 
22 


53 


114.  Case  IV: 


5'  26" 
22      7 
42    34 

5=   42C 
c  =  153 
(7=160 
#  =  137 
c'=   90 
<7'  =   50 

Given  A,  B  and  a,  to  Find  c  and  &. — 
tan  <f>  =cosB  tan  a, 
sin  $'= sin  <£  cot  A  tan  #, 
cos  b  =  cos  a  cos  <£'  sec  <£, 

c = <p  -\-  <$>'  and 
is  determined  by  its 
sine,  it  may  be  in  either  the  first 
or  the  second  quadrant;  and  c, 
therefore,  has  two  values,  as  noted. 
The  corresponding  two  values  of 
b  are  determined  from  the  two 
values  of  4,'. 

The  angle,  C,  if  required,  is  determined  from  the  formulae : 

cot  6  =  cos  a  tan  B,  (5) 

cot  &  -  cot  6  tan  <f>  cot  </>',  ( 6 ) 

C=(0+tf)  or  0+(180°-0').  (7) 

This  problem  does  not  occur  in  navigation,  and  is  of  rare  occur- 
rence elsewhere. 

8 


Since 


98  SPHERICAL  TRIGONOMETRY. 

Examples. 

1.  Given  4  =  115°  36'  45" 

B=   80     19    12 
6=   84     21    56     find  a,  c,  and  (7. 

Ans.     0=114°  26'  50" 
c=   82     33    31 
C=   79     10    30 

2.  Given  a=53°   18'  20" 

5  =  46     15    15 

4  =  79     30    45     find  &,  c,  and  (7. 

Ans.     c  =  50°  24'  57" 
(7=70     55    35 
&  =  36       5    34 

3.  Given  B  =   76°  41'  13" 

4  =  117     44    36 
a=126     17    22     find  I,  c,  and  C. 

Ans.     6  =  117°  35'  35" 
c=   24     16    50 
C=   26     50    24 

&'  =  62  24  25 
c'  =  120  53  50 
(7  =  109  34  34 


115.  Case  V:    Given  the  Three  Sides,  a,  b  and  c  — 


4t  this  problem  is  most  conveniently  and  accurately  solved  by 
means  of  the  tangents  of  the  half  -angles  (Art.  W-),  especially 
when  all  three  angles  are  required. 

It  is  one  of  the  most  frequent  problems  of  navigation  (where, 
however,  only  the  angles  4  and  B  are  required)  and,  for  reasons 
peculiar  to  the  problem,  4  is  found  from  sin2  £4  (Art.  88),  and 
B  from  cos2£#  (Art.  89). 

The  tangent  formulae  may  be  put  in  a  more  convenient  form  by 
multiplying  the  expressions  of  Art.  90  by 

sin(s  —  a]      sin(s—  b)        -,  sin(s  —  c) 
sin(s  —  a)  '    sin(s  —  b)  sin(s  —  c}  ' 


SPHERICAL  TRIGONOMETRY.  99 

respectively,  giving 

tan   B= 


,  —   ~ 

sm(s-a)  sm(s-b) 

P 


sin (5  —  c)  9 
where 

p—     /sin  (s  —  a]  sin  (s  —  b )  sin  (s— c) 
V  sins 

Example:    Given  a=56°  37',  6  =  108°  14',  c=75°  29'. 


a   =  56 
&   =108 
c   =75 

37 
14 
29 

0 
0 
0, 

cosec  0.30933    0  .  0  6  33.0 

sin   9.95198 
sin   9.31549 
sin   9.84707 

s   =120 

10 

0 

5—6= 

63 
11 

44 

33 
56 
41 

0 
0 
0 

P  .  . 

2)9.17774 

9.58887 

U  = 
i*  = 

io  = 

23 

61 

28 

25 
56 
53 

55 

54 

28 

tan 
tan 
tan 

9.63689 
0.27338 
9.74180 

These  results  should  be  checked  by  the  sine  formula. 


100  STEREOGRAPHIC  PROJECTIONS. 


STEREOGRAPHIC  PROJECTIONS. 

THE  FOLLOWING  CHAPTERS,  TOGETHER  WITH  THE  PRECEDING  METHODS  OF  SPHER- 
ICAL TRIGONOMETRY  ARE  ESSENTIAL  TO  THE  STUDY  OF  NAVIGATION;  AND, 
AS  A  PROFESSIONAL  SUBJECT,  THEY  SHOULD  BE  MOST  THOROUGHLY  UNDER- 
STOOD BY  ALL  MIDSHIPMEN. 


CHAPTEE  XII. 

116.  Definitions. — The  projection  of  the  sphere  on  the  plane  of 
one  of  its  great  circles,  when  the  point  of  sight  is  taken  at  one  of 
the  poles  of  the  great  circle  on  which  the  projection  is  to  be  made, 
is  called  a  stereo  graphic  projection. 

The  plane  on  which  the  projection  is  made  is  called  the  primi- 
tive plane,  and  the  circle  the  primitive  circle. 

The  polar  distance  of  a  point  on  the  sphere  is  its  angular  dis- 
tance on  the  surface  of  the  sphere  from  one  of  the  poles  of  the 
primitive  circle. 

The  polar  distance  of  a  circle  is  the  angular  distance  of  any 
point  of  its  circumference  from  either  of  its  own  poles. 

117.  The  inclination  of  a  circle  is  the  angle  between  its  plane 
and  the  primitive  plane. 

Let  WBED  (Fig.  59)  be  a  great  circle  cut  from  the  sphere  by 
the  primitive  plane,  N  and  8  its  poles,  the  pole  8  being  taken  as 
the  point  of  sight. 

Let  P!  be  any  point  of  the  sphere,  and  NP^BS  a  great  circle 
cut  from  the  sphere  by  a  plane  through  P±  intersecting  the  primi- 
tive plane  in  the  straight  line  BOD.  Then  NPt  is  the  polar  dis- 
tance of  P1 ;  and  ply  the  point  in  which  SP1  pierces  the  primitive, 
is  the  stereographic  projection  of  the  point  Px. 


STEREOGRAPHIC  PROJECTIONS', 


101 


The  angle  NSP^  is  measured  by  one-half  the  polar  distance  of 
Pl ;  hence  Op^  the  distance  of  p±  from  the  center  of  the  primitive 
circle,  is 

Op^-rtan^p, 

where  p  is  the  polar  distance,  and  r  the  radius  of  the  primitive 
circle. 

N 


^  .— 

\                          1                   S~~  -~ 

!    \     !//'      "^ 

\         i7^/ 

A.           j^: 

< 

.\      /<|0 

X.                        ^ 

\l 


118.  Let  Cd-R  (Fig.  59)  be  a  small  circle  of  the  sphere,  its 
plane  making  an  angle  <j>  with  the  primitive  plane.  Its  pole  P 
is  the  extremity  of  the  diameter  of  the  sphere,  OP,  passing 
through  the  center  of  the  circle  perpendicular  to  its  plane. 

The -polar  distance  of  the  circle  is  PQ  —  PR.  Its  inclination, 
(f>,  to  the  primitive  WEED  is  measured  by  the  arc  NP,  which 
measures  the  angle  between  their  poles,  since  the  angle  between 
the  planes  is  equal  to  the  angle  between  the  normals  ON  and  OP 
drawn  from  0  to  the  two  planes. 


102 


STEREOGPA^HIC  PROJECTIONS. 


The  line  WE,  the  intersection  of  the  plane  passed  through  NO 
and  OP  with  the  primitive  plane,  is  called  the  line  of  measures  of 
the  circle  QQ^R. 

119.  Let  NSWE  (Fig.  60)  be  a  circle  cut  from  the  sphere  by  a 
plane  through  NS  and  OP,  the  axes  of  the  primitive  circle  and  the 
small  circle  QR,  respectively,  and  WE  the  straight  line  in  which 
it  intersects  the  primitive  plane.  Then  WE  is  the  line  of  measures 
of  the  circle  QR. 

N 


P  being  the  pole  of  the  circle  QR,  PQ  =  PR  is  its  polar  distance, 
and  PN  its  inclination  (since  the  angle  between  two  circles  is 
equal  to  the  distance  between  their  poles) .  8  is  the  point  of  sight, 
and  q  and  r  the  projections  of  the  extreme  or  principal  elements 
of  the  oblique  circular  cone  SQR,  which  is  formed  by  the  pro- 
jecting lines  of  all  points  of  the  circle  QR. 

Denoting  the  polar  distance  of  the  circle  QR  by  p,  and  its 
inclination  by  <£,  we  have,  in  the  two  right  triangles  SOq  and  SOr 


STEEEOGRAPHIC  PROJECTIONS. 


103 


-at"  the  distances  on  the  line  of  measures  from  the  center  of  the 
primitive  circle  to  the  projected  extremities  of  the  diameter  of  QR. 

120.  The  Stereographic  Projection  of  a  Circle  is  a  Circle. — 
Let  QR  (Fig.  61)  be  any  circle,  P  the  vertex  of  a  cone  tangent  to 


the  sphere  along  QR,  and  Q  any  point  in  QR.  Pass  a  plane 
through  SQP,  cutting  from  the  primitive  a  line,  pq,  which  is  the 
projection  of  the  element  PQ;  cutting  a  tangent  plane  at  8  in  ST, 
which  is  parallel  to  pq;  and  from  the  sphere  a  circle  (not  shown 
in  the  figure)  to  which  PQ  and  ST  are  tangents,  8Q  being  the 
chord  of  contact.  Fig.  62  represents  the  circle  omitted  in  Fig.  61, 
rotated  into  the  plane  of  the  paper. 


104  STEREOGRAPHIC  PROJECTIONS. 

Draw  PK  parallel  to  pq  (Fig.  62)  ;  it  is  also  parallel  to  TS, 
and  the  alternate  interior  angles  TSQ  and  QKP  are  equal.  But 
as  T8  and  TQ  are  tangents  from  a  common  point  T,  they  are 
equal,  and  the  angle  TSQ  =  angle  TQS Bangle  KQP.  Hence 
PK=PQ. 

K-, .,? 


FIG.  62. 


The  triangles  Sqp  and  SKP  are  similar,  hence 

or  pq  =  PQ  -        ,  which  is  equal  to  a  constant,  since  PQ, 


pS  and  PS  are  all  fixed  lengths. 

Since  pq  is  constant,  the  point  p  being  fixed,  it  follows  that  the 
locus  of  q  is  a  circle  whose  center  is  the  projection  of  the  vertex 
of  the  tangent  cone. 


STEREOGRAPH: c  PROJECTIONS. 


105 


We  shall  see  hereafter  that  this  gives  us  a  simple  construction, 
to  find  the  center  of  the  projected  circle. 

121.  Angles  on  the  Sphere  are  Unaltered  in  the  Stereographic 
Projection.— Let  MR  and  MT  (Fig.  63)  be  tangents  to  two 
curves  on  the  sphere  at  the  common  point  M.  (These  curves  are 


S  "*--.. 


FIG.  63.  T 

not  shown  in  the  figure.)  Let  these  tangents  be  projected  on  the 
primitive  plane  by  the  planes  RIMS  and  TMS,  respectively,  in  the 
lines  mr  and  mi,  and  let  MaS  and  Mis  be  the  circles  cut  from  the 
sphere  by  these  planes;  also  let  the  lines  cut  from  the  tangent 
plane  at  8  be  SR  and  ST.  Since  the  tangent  plane  is  parallel  to 
the  primitive,  the  lines  SR  and  ST  will  be  parallel  to  mr  and  mt, 
and  the  angle  RST-rmt. 


106 


STEREOGRAPHIC  PROJECTIONS. 


Join  RT.  Now,  since  R8  and  RM  are  tangents  to  the  same 
circle,  they  are  equal;  and  for  the  same  reason  TM=TS;  hence 
the  triangles  RMT  and  R8T  are  equal,  and  the  angle  RMT= 
RST=rmt.  That  is,  the  angle  between  two  lines  drawn  tangent 
to  the  sphere  at  a  common  point  is  equal  to  the  angle  between  their 
projections. 

Y 


FIG.  64. 

/ 

As  the  angle  between  two  curves  on  the  sphere  is  measured  by 
the  angle  between  the  tangents  at  the  common  point,  it  follows 
that  angles  on  the  sphere  are  unaltered  by  'the  stereographic  pro- 
jection. 

122.  The  principal  properties  of  the  stereographic  projection 
are  those  just  proved;  viz.: 

1.  That  all  circles  are  projected  as  circles. 

2.  That  angles  are  not  altered. 


STKREOGRAPHIC  PROJECTIONS. 


107 


These  properties  enable  us  to  construct  the  projections  of 
spherical  triangles  by  convenient  and  simple  methods. 

123.  To  Project  Any  Circle  of  the  Sphere,  Given  its  Polar  Dis- 
tance and  the  Projection  of  its  Pole. — First  Method. — Let  NESW 


be  the  primitive  circle  (Fig.  64),  and  Z  the  projection  of  the 
point  diametrically  opposite  the  point  of  sight,  the  latter  lying 
in  the  axis  of  the  primitive  circle  at  a  perpendicular  distance  r 
from  Z,  below  the  primitive  plane. 

Let  P  be  the  projection  of  the  pole  of  the  given  circle ;  draw  the 
diameters  NS  through  P,  and  WE  at  right  angles.  The  diameter 
NPS  is  the  projection  of  the  great  circle  through  P;  and  we  re- 
volve it  about  the  diameter  N8  into  the  plane  of  the  primitive, 
bringing  tlie  point  of  sight  to  W.  NS  is  the  line  of  measures,  and 


108 


STEREOGRAPH: c  PROJECTIONS. 


Q}  the  intersection  of  WP  (produced]  with  NQE,  is  the  revolved 
position  of  the  pole  P. 

Lay  off  from  Q,  Qr  and  Qt  equal  to  the  given  polar  distance, 
and  draw  Wr  and  Wt,  cutting  the  line  of  measures  in  r±  and  t19 
the  extremities  of  the  diameter  of  the  required  projection  (Art. 
119),  and  c,  the  middle  point  of  r±£1?  is  its  center. 

f 


Second  Method. — Find  Q,  as  in  the  first  method,  take  Qr  (Fig. 
64)  equal  to  the  given  polar  distance,  and  draw  Wrjr,  giving  one 
extremity  of  the  diameter  at  rv.  Draw  rV  tangent  at  r,  meeting 
ZQV  in  V,  the  vertex  of  the  tangent  cone,  and  join  VW,  cutting 
NS  in  c,  the  center  of  the  required  circle  (see  Art.  120,  Fig.  61). 
Draw  the  circle  with  radius  c/v 

This  method  is  most  convenient  when  the  pole  is  on  the  circum- 
ference of  the  primitive  circle,  as  in  Fig.  65.  Here  we  simply 
draw  the  diameter  NS  through  P  and  WE  at  right  angles,  laying 
off  the  polar  distance  Pr,  and  drawing  the  tangent  at  r  to  intersect 
NS  in  the  center  c. 


STEREOGRAPHIC  PROJECTIONS.  109 

124.  To  Project  a  Great  Circle. — When  the  polar  distance  is 
90°,  the  circle  is  a  great  circle  and  passes  through  W  and  E  (Fig. 
66),  which  are  each  90°  from  P. 

Proceed  as  for  a  small  circle  to  find  Q,  lay  off  0r  =  90°,  and 
draw  Wrtr.  The  tangent  cone  to  the  sphere  becomes  in  this  case 
a  tangent  cylinder,  of  which  ZQ  is  the  axis ;  and  the  line  from  W, 
parallel  to  ZQ,  cuts  the  line  of  measures  at  c,  the  required  center. 
Draw  the  circle  with  radius  cr^  —  cW. 

Since  W,  r±  and  E  are  three  points  of  the  required  circle,  a  per- 
pendicular to  the  middle  point  of  Wr^  will  intersect  N8  in  c,  a 
much  less  convenient  way  of  getting  c. 

125.  To  Find  the  Locus  of  the  Centers  of  the  Projections  of 
All  Great  Circles  Passing  Through  a  Given  Point.— Let  P  (Fig. 
67)  be  the  given  point  through  which  the  projections  of  great 
circles  are  to  pass;  draw  the  diameters  NFS  and  WE  at  right 
angles,  and  draw  WPQ  to  Q. 

1.  The  projections  of  all  great  circles  through  P  must  also  pass 
through  a  point  180°  from  P;  hence  draw  the  diameter  Qr,  and 
draw  Wr,  cutting  NS,  the  line  of  measures,  in  T;  then  T  is  the 
projection  of  the  point  180°  from  P. 

Since  all  the  required  circles  pass  through  P  and  Tf  their  center 
must  lie  on  the  straight  line  perpendicular  to  PT  at  its  middle 
point  c;  the  line  is  called  the  line  of  centers. 

2.  Since  a  great  circle  may  always  be  drawn  through  the  points 
W,  P  and  E  (Art.  124),  the  point  c  may  be  found  by  drawing  a 
perpendicular  to  WP  at  its  middle  point,  intersecting  N8  in  c. 

3.  The  triangle  WcP  is  isosceles,  and  the  angle  PWs=ihe  angle 
WPS,  which  is  measured  by  J(90°  +QN)  =%QNW;  i.  e.,  the  arc 
QEs  =  QNW.    Hence  lay  off  QEs=QNW,  and  draw  Wcs.    This 
is  equivalent  to  laying  off  a  polar  distance  QNW  from  Q;  and 
thus  the  line  of  centers  is  the  projection  of  a  small  circle  passing 
through  the  iwe  of  sight  and  having  the  polar  distance  QNW 
=  180°— <£,  where  <f>  denotes  its  inclination. 


110 


STEREOGRAPHIC  PROJECTIONS. 


4.  From   the   figure,    Wr  =  QE,   and   rSs=lSO° -QEs  =  180° 
-QNW=QE.    Hence  lay  oft*WSs=2QEf  and  draw  Wcs. 

5.  Since  the  angle  WEs=QZE=WZr,  a  line  joining  E  and  s 
is  parallel  to  Qr,  which  gives  a  fifth  method. 

Of  these  methods  the  most  convenient  are : 

( 3 )  Lay  off  QEs  =QNW  and  draw  Wsf 

(4)  l&joftWS8=2QE. 


126.  To  Draw  a  Great  Circle  Through  P  Making  a  Given  Angle 
with  NS.— The  tangent  to  the  required  circle  at  P  (Fig.  67) 
makes  the  required  angle  (x)  with  PZS  (Art.  121) ;  the  per- 
pendicular to  the  tangent  (i.  e.,  the  radius),  makes  with  PZS  the 
angle  90°—  x.  Hence  construct  ZPC  =  9Q°—x,  intersecting  the 
line  of  centers  in  C,  the  center  of  the  required  circle. 


STEREOGRAPHIC  PROJECTIONS. 


Ill 


Note  that  the  projection  of  a  great  circle  always  meets  the 
primitive  circle  at  the  extremities  of  a  diameter,  LK,  of  the 
primitive  circle. 

127.  To  Find  the  Pole  of  a  Given  Circle.— 1.  Let  Wr^E  (see 
Fig.  66)  be  a  given  great  circle.  Draw  the  diameters  WE  and  NS 
at  right  angles,  and  draw  Wr±r.  Lay  off  rQ  =  9Q°  and  join  WQ, 
cutting  NS  in  P,  the  required  pole. 


2.  Let  r^  (Fig.  64)  be  a  given  small  circle;  through  its  center 
c  draw  the  diameter  NS,  and  WE  at  right  angles.  Draw  Wr^  and 
Wt19  intersecting  the  primitive  circle  in  r  and  t.  Bisect  the  arc 
tQr  in  Q,  and  draw  WQ  intersecting  NS  in  P,  the  required  pole. 

128.  Given  the  Projected  Arc  of  a  Great  Circle,  to  Find  its 
True  Length. — Let  NM  (Fig.  68)  be  the  arc  to  be  measured. 
Fing  the  pole  X  of  this  arc,  and  draw  the  straight  line  XMm; 
then  Nm  is  the  true  length  required.  For  XMm  is  the  projection 


112  STEREOGRAPHIC  PROJECTIONS. 

of  a  small  circle  which  passes  through  the  point  of  sight;  i.  e., 
through  a  pole  of  the  primitive ;  and  since  it  passes  through  the 
poles  of  two  circles,  NM  and  the  primitive,  it  makes  equal  angles 
with  them,  the  triangle  NMm  is  isosceles,  and  NM=Nm. 

The  distance  Nm  might  also  be  found  by  drawing  the  tangent 
Me  to  meet  NS  at  c;  then  a  tangent  to  the  primitive  through  c 
meets  the  primitive  at  ra.  In  other  words,  c  is  the  vertex  of  a 
tangent  cone  of  which  a  small  circle  passing  through  M  and  m 
(pole  at  N)  is  the  base.  But  this  method  would  generally  be  in- 
convenient. 

The  solution  of  the  reverse  problem  is  obvious,  to  lay  off  the  pro- 
jected length  NM  on  a  given  circle  Nr^S,  from  its  true  length  Nm, 
by  simply  drawing  from  the  pole  X,  Xm  to  m,  cutting  the  given 
circle  in  M . 

129.  Spherical  Triangles  on  the  Earth  Regarded  as  a  Sphere. 
Definitions. — The  axis  of  the  earth  is  the  diameter  about  which 
the  earth  revolves. 

The  north  and  south  poles  are  the  extremities  of  the  earth's  axis. 

The  equator  is  a  great  circle  of  the  earth  perpendicular  to  its 
axis. 

Meridians  are  great  circles  of  the  earth  passing  through  its 
poles,  and  are  therefore  perpendicular  to  the  equator. 

The  latitude  of  a  point  on  the  earth's  surface  is  its  angular  dis- 
tance north  or  south  of  the  equator  (measured  on  the  meridian 
passing  through  the  point) . 

The  longitude  of  a  point  on  the  earth's  surface  is  the  angle  at 
the  poles  between  the  meridian  of  the  point,  and  a  fixed  meridian 
taken  as  the  origin  of  longitudes,  as  the  meridian  of  Greenwich, 
or  of  Washington  Longitude  is  reckoned  from  the  zero  meridian, 
positive  to  the  west,  negative  to  the  east,  from  Oh  to  12h,  or  from 
0°  to  180°. 

130.  To  Find  the  Great  Circle  Distance  between  Two  Given 
Points. — The  angle  between  the  meridians  of  the  two  planes  is 


STEREOGRAPHIC  PROJECTIONS. 


113 


obviously  the  algebraic  difference  of  their  longitudes.  The  two 
adjacent  sides  are  the  co-latitudes  of  the  two  places ;  and  we  have 
two  sides  and  the  included  angle  (Fig.  69)  to  project  the  triangle, 
measure,  and  compute  the  desired  parts. 


FIG.  69. 


Example  :  Find  the  great  circle  distance  from  Cape  Flattery, 
Long.  (AJ  =8h  20m  W.,  Lat.  (LJ  =48°  N.,  to  Java,  Long.  (X2) 
=  7h  28mE.,  Lat.  L2  =  9°  S. 


Take  the  meridian  of  Cape  Flattery  as  the  primitive  circle  (Fig. 
69)  and  P  as  the  North  Pole  of  the  earth.    Lay  off  PM1  =  90°  —L^ 
9 


114 


STEREOGRAPHIC  PROJECTIONS. 


=  42°.  Draw  through  P  a  great  circle  making  the  angle  123° 
with  PMi  (or  33°  with  PP')  .  See  Art.  126.  This  is  conveniently 
done  by  laying  off  P'Q'c—  114°,  twice  the  angle  cfP',  then  draw- 
ing PCC-L  to  c±  the  center  of  the  required  circle  PM2P'.  Find  its 
pole  p0  by  Art.  127.  Make  PQs=99°  (the  north  polar  distance 
of  Java,  9°  S.),  and  draw  p0s  cutting  off  PM2,  the  projected  length 
of  99°.  Draw  the  diameter  M^m^  and  the  diameter  perpendicular 
to  it;  then  find  the  center  of  the  great  circle  through  MJlz  by 
perpendicular  bisector  of  chord  M^MZ)  meeting  line  of  centers  at 
c2,  and  draw  MJttzm^  The  triangle  PM^M2  is  the  projection  of 
the  given  triangle. 

After  finding  the  pole,  p2,  of  M^M2m^  the  true  length  of  M^M., 
is  given  on  the  primitive  circle  (Art.  128)  by  the  arc  D  intercepted 
between  p^M^  and  p2M2  (produced),  about  118°  30'  .  The  angle 
PM^M2  is  measured  by  the  projected  arc  mp  of  the  great  circle 
90°  from  M19  the  true  length  of  which  is  marked  M^  on  the  primi- 
tive circle,  about  70°. 

131.  Numerical  Computation.  —  From  Art.  105,  Spherical 
Trigonometry,  calling 

\2-Xi=A,     6  =  90°-L2,     c=90°-L1, 
we  get 

tan  (f>  =  cos  (  A2  —  Aj  )  cot  L2, 

$'  =  (90°  -LJ  -$  = 
cosZ>  =  sinL2  sec  <£  sin( 
cot(A2  —  Ajcosec 


L!  48°  00'  00" 

X2— \,=123°  00'  00" 
^2 


4 
— _9°  00'  00" 

=  73°  47'     8" 


cos 
cot 
tan 


9.7361  In 
0.80029w 


cot     9.81252n 


D 

M, 

D 


0.53640 
47'    8" 

=118°  30'  35" 
=  70°  23'  30" 
=7110.6  geographical  miles. 


sin  9.19433w 

sec  0.55404 

sin  9.92943 

cos  9.67880w 


esc 
cos 


0.01763 
9.72160n 


cot     9.55175 


STBREOGRAPHIC  PROJECTIONS.  115 

The  values  found  from  the  projection  agree  as  closely  with  the 
computed  values  as  could  be  expected  from  the  scale  of  the  pro- 
jections (Fig.  69). 

Examples. 

Project  the  triangle  and  compute  the  great  circle  course 
between : 

(1)  San  Francisco,      L1  =  37°  47'  48"  N.,  A1  =  8h  9ra  43s  W. ; 
Manila,  L2  =  14°  35'  25"  N.,  A^=8h  3m  50s  E. 

Ans.     0  =  100° '50'  55". 

M^N.  61°  45'  34"  W. 

(2)  New  York,  L=40°  40'  N.,  A=   4h  55m  54s  AY.; 
Cape  Good  Hope,        L  =  33°  56'   S.,  X=   lh  13m  55s  E. 

(3)  San  Francisco,  L  =  37°  47'  N.,  X=   8h  09m  43s  W.; 
Sydney,  L  =  33°  52'  S.,  A=10h  04m  50s  E. 

132.  The  spherical  coordinates  of  a  point  on  the  earth,  latitude 
and  longitude,  as  defined  in  Art.  129,  can  be  determined  only  by 
astronomical  observations  of  the  heavenly  bodies,  the  sun,  moon, 
planets  and  fixed  stars.  ^  \    (fc^ftf 

^In  such  observations  these  01103  cotes  are  seen  projected  on  the 
background  of  the  sky,  and  appear  to  have  a  revolution  about  an 
axis  from  east  to  west.  This  apparent  revolution  is  due  to  the 
rotation  of  the  earth  on  its  axis  in  the  opposite  direction,  but  we 
shall  regard  the  apparent  revolution  of  the  heavenly  bodies  as  real. 

133.  The  Celestial  Sphere. — The  celestial  sphere  is  an  imagi- 
nary spherical  surface  of  indefinite  radius,  having  the  eye  of  the 
observer  at  its  center,  and  upon  which  the  heavenly  bodies  are 
projected. 

The  poles  of  the  sphere  are  the  points  in  which  the  earth's  axis 
produced  meets  the  celestial  sphere,  and  about  which  the  celestial 
sphere  apparently  revolves  once  in  24  hours. 

The  equinoctial  (or  celestial  equator)  is  the  great  circle  90° 
from  the  poles — obviously  the  same  as  the  great  circle  formed  by 
the  indefinite  extension  of  the  earth's  equator. 


116  STEREOGRAPHIC  PROJECTIONS. 

Hour  circles  are  great  circles  through  the  poles — or  the  great 
circles  produced  by  the  indefinite  extension  of  the  earth's 
meridians. 

The  zenith  and  nadir  are  the  points  of  the  celestial  sphere 
vertically  overhead  and  vertically  underneath  the  observer.  They 
may  be  conceived  to  be  the  points  in  which  the  celestial  sphere  is 
pierced  by  the  indefinite  extension  of  the  radius  of  the  terrestrial 
sphere  -through  the  position  of  the  observer. 

The  celestial  horizon  is  the  great  circle  of  which  the  zenith  and 
nadir  are  the  poles ;  or  it  is  the  great  circle  cut  from  the  celestial 
sphere  by  the  tangent  plane  to  the  earth  at  the  observer's  position. 

Since  the  radius  of  the  celestial  sphere  is  infinite,  we  may  sub- 
stitute for  the  tangent  plane,  a  parallel  plane  through  the  center 
of  the  earth,  for  these  two  planes  will  meet  the  celestial  sphere 
in  the  same  great  circle,  which  is  the  celestial  horizon.  (See  Fig. 
71,  Art.  135.) 

Vertical  circles  are  great  circles  passing  through  the  zenith  and 
nadir. 

The  celestial  meridian  (of  a  place)  is  the  hour  circle  passing 
through  the  zenith  and  nadir ;  it  may  also  be  defined  as  the  vertical 
circle  passing  through  the  poles.  The  celestial  meridian  is  ob- 
viously the  great  circle  formed  by  the  indefinite  extension  of  the 
plane  of  the  terrestrial  meridian  of  a  given  place. 

The  prime  vertical  is  the  vertical  circle  passing  through  the  east 
and  west  points  of  the  horizon,  and  is  perpendicular  to  the 
meridian  and  to  the  horizon. 

The  declination  (d)  of  a  point  of  the  celestial  sphere  is  its 
angular  distance  north  or  south  of  the  equinoctial,  measured  on 
the  hour  circle  passing  through  the  point.  It  is  marked  north  or 
south  like  latitude  on  the  earth. 

The  polar  distance  (p)  of  a  point  of  the  celestial  sphere  is  its 
angular  distance  from  the  pole;  p  =  9Q°  —  d  if  the  declination  is  of 


STEREOGRAPHIC  PROJECTIONS.  117 

the  same  name  as  the  pole  from  which  it  is  reckoned,  otherwise 
p  =  90°  +  d. 

The  altitude  (h)  of  a  body  is  its  angular  distance  above  the 
horizon. 

The  zenith  distance  (z)  of  a  body  is  its  angular  distance  from 
the  zenith;  z  =  90° -h. 

The  hour  angle  (t)  of  a  point  of  the  celestial  sphere  is  the  angle 
at  the  pole  between  the  meridian  of  a  place  and  the  hour  circle 
passing  through  the  point. 

The  azimuth  (Z)  of  a  body  is  the  angle  at  the  zenith  between 
the  meridian  of  a  place  and  the  vertical  circle  passing  through  the 
body. 

The  amplitude  of  a  body  is  the  arc  of  the  horizon  between  the 
east  or  west  point  and  the  body  when  in  the  horizon. 

The  ecliptic  is  the  great  circle  of  the  celestial  sphere  which  is 
the  apparent  path  of  the  sun,  due  to  the  real  motion  of  the  earth 
around  the  sun. 

The  vernal  equinox,  or  the  first  point  of  Aries,  is  the  point  in 
which  the  ecliptic  cuts  the  equinoctial,  as  the  sun  passes  from 
south  tojnorth  declination. 

The  right  ascension  of  a  heavenly  body  is  the  angle  at  the  pole 
between  the  hour  circle  passing  through  the  body  and  the  hour 
circle  through  the  vernal  equinox. 

134.  The  Astronomical  Triangle. — The  spherical  triangle 
whose  vertices  are  the  pole,  the  zenith  and  any  point  M  on  the 
celestial  sphere  is  called  the  astronomical  triangle.  The  angle  at 
the  pole  P  is  the  hour  angle  (t),  the  angle  at  Z  is  the  azimuth 
(Z),  the  angle  at  the  point  M  is  the  position  angle.  The  side 
PZ  is  the  co-latitude  (90°  -L) ;  the  side  PM  is  the  polar  distance 
from  the  elevated  pole,  and  is  equal  to  90°  — d  when  the  latitude 
and  declination  have  the  same  name,  to  90° +d  when  they  have 
different  names ;  the  side  ZM  is  the  zenith  distance  of  the  point, 
or  the  co-altitude,  90°—  h. 


118 


STEREOGRAPH ic  PROJECTIONS. 


The  relation  of  the  astronomical  triangle  to  the  definitions  of 
Art.  133  is  shown  by  the  diagram  of  Fig.  70 : 


NWS  =  Horizon. 
Mrrii  —  Vertical  circle  of  M. 
Mm,  =  Altitude  of  M. 
=  Azimuth  of  M. 


PQP'Q'  =  Meridian. 
(Q'WQ  =  Equator. 
I  PMP'  =  Hour  circle  of  M . 
mM  —  Declination  of  M . 
[  MPZ  =  Hom  angle  of  M. 

Either  of  these  two  groups  forms  &'  system  of  spherical  co- 
ordinates, by  which  the  position  of  a  body  on  the  celestial  sphere 
is  known  with  reference  to  the  observer's  position : 

1.  When  its  hour  angle  and  declination  (or  polar  distance)  are 
known. 

2.  When  its  azimuth  and  altitude   (or  zenith  distance)   are 
known. 

135.  The  latitude  of  a  place  on  the  earth  is  equal  to  the  declina- 
tion of  its  zenith  or  to  the  altitude  of  the  elevated  pole. 


STEREOGRAPHIC  PROJECTIONS. 


119 


Let  pzq  (Fig.  71)  be  the  terrestrial  meridian  of  a  place  z,  pp^ 
the  axis  of  the  earth,  and  q^Oq  the  equator. 

Extend  the  axis  of  the  earth,  the  equator,  and  the  radius  (or 
vertical)  of  the  place,  to  meet  the  celestial  sphere  in  P,  Q  and  Z. 
The  plane  NOS,  parallel  to  the  tangent  plane  nzs  at  z,  is  the 
celestial  horizon. 

The  arc  qz,  which  is  by  definition  the  latitude  of  z,  is  obviously 
equal  to  QZ,  which  by  definition  is  the  declination  of  the  point  Z, 
the  zenith. 


FIG.  71. 

The  elevated  pole  (P)  is  the  pole  visible  above  the  observer's 
horizon  (the  north  pole  for  places  in  north  latitude,  the  south  pole 
for  places  in  south  latitude) . 

The  arc  NP,  the  altitude  of  the  elevated  pole  P,  is  equal  to  QZ, 
the  latitude  of  z,  since  each  is  equal  to  90°  less  the  arc  PZ.  The 
arc  PZ,  included  between  the  zenith  and  the  pole,  is  therefore 
90°  —L,  where  L  is  the  latitude  of  a  place. 

136.  To  Find  the  Latitude  of  a  Given  Place.— The  latitude  of 
a  given  place  (i.  e.,  the  declination  of  its  zenith)  is  most  easily 
found  from  the  observation  of  the  altitude  of  the  sun  or  other 


120 


STEREOGRAPHIC  PROJECTIONS. 


celestial  body  when  it  is  just  on  the  meridian  of  the  place.  Sup- 
pose the  meridian  altitude  of  such  a  body  to  have  been  measured 
and  reduced  by  various  small  corrections  to  give  the  true  altitude 
above  the  celestial  horizon.  The  declinations  of  the  brighter 
celestial  bodies  are  tabulated  in  the  American  Nautical  Almanac. 
With  these  two  data  the  method  of  finding  the  latitude  is  shown 
in  Fig.  72,  where  the  celestial  sphere  is  projected  on  the  plane  of 


the  meridian  of  the  given  place.  ZSNaN  is  the  meridian,  Z  the 
zenith,  NWS  the  horizon,  P  the  elevated  pole,  Q^Q  the  equator, 
and  QZ  the  latitude.  If  m±  is  the  observed  body,  then  QM±  is  its 
declination  and  m^Z  its  zenith  distance  =  90°  —  h:.  We  have  from 
the  figure 


If  the  body  is  in  south  declination,  as  at  ra2,  then 
L=ZMz-QM2=Z2+(-d2), 


STEREOGRAPHIC  PROJECTIONS. 


121 


south  declination  in  this  case  being  negative  (the  latitude  being 
north). 

If  the  body  is  north  of  the  zenith,  as  at  M3,  then 


FIG.  73. 

137.  Projection  of  the  Astronomical  Triangle.    Case  I :    Given 

L  =  20°  N.,  d=32°  N.,  h  =  4:Q°,  body  bearing  W.,  to  Project  the 
Triangle  on  the  Plane  of  the  Horizon. — The  three  sides  of  the 
given  triangle  are  PZ  =  70°,  PM=58°,  ZM  =  50°. 

Let  NE8W  be  the  horizon,  then  Z  will  be  the  center  of  the 
primitive  circle  (Fig.  73).    Draw  the  diameters  NZ8  and  WZE 


122  STEREOGRAPHIC  PROJECTIONS. 

at  right  angles,  and  take  NZ8  as  the  line  of  measures.  The  point 
of  sight  (Na)  lies  below  Z  perpendicular  to  the  primitive  plane 
and  at  a  distance  r.  The  projected  pole  P  lies  between  Z  and  Nf 
and  for  the  purpose  of  finding  this,  the  point  of  sight  is  revolved 
about  NS  into  the  position  W. 

Lay  off  EA  =  90°  -  L,  and  draw  WA  cutting  N8  in  P,  the  pro- 
jected pole.  Construct  the  small  circle  of  58°  polar  distance  from 
P  by  laying  off  Ar  and  Ar'  equal  to  the  polar  distance  58°,  and 
joining  Wrf  intersecting  NS  in  r±  one  extremity  of  the  diameter 
(Art.  123).  The  center  c  is  found  by  the  second  method  of  Art. 
123,  the  tangent  at  /  intersecting  ZA  (produced)  in  V,  and  join- 
ing WV,  cutting  NS  in  c;  cr^  is  the  radius  of  the  required  circle. 

Lay  off  EH=9Q0-h  =  5Q°,  and  draw  WH  cutting  N8  in  h; 
then  Zh  is  the  polar  distance  50°  from  Z.  With  Zh  as  a  radius 
(and  Z  as  center)  draw  a  circle  cutting  Mr±  in  M.  This  point 
being  at  the  given  distance  from  both  P  and  Z  respectively,  and 
lying  west  of  the  meridian  NS,  is  the  required  point. 

Lay  off  As=ANW,  and  join  Ws>  cutting  NS  in  c±;  draw  the 
line  of  centers  of  great  circles  through  P,  perpendicular  to  NS 
at  c±.  Bisect  the  chord  PM,  and  draw  a  perpendicular  at  its 
middle  point  meeting  the  line  of  centers  in  c,  the  center  of  the 
required  great  circle  PM. 

Draw  the  great  circle  ZM,  passing  through  the  point  of  sight, 
as  a  straight  line,  completing  the  construction  of  PZM.  The 
values  of  the  required  quantities  t  and  Z  are  indicated  on  the 
projection ;  the  measure  of  the  angle  Z  by  the  arc  of  the  primitive 
circle  included  between  ZPN  and  ZM  produced,  approximately 
64° ;  the  value  of  the  hour  angle  ZPM=t  is  measured  by  the  arc 
of  the  primitive  circle  marked  t,  approximately  53°  45',  which 
is  the  true  length  of  Qm  the  projected  arc  of  the  equator,  WQE, 
intercepted  between  the  two  hour  circles  PZQ  and  PMm.  The 
angle  PcCl  is  also  a  measure  of  the  angle  ZPN.  (See  Art.  126.) 


STEREOGRAPHIC  PROJECTIONS. 


123 


138.  To  Project  the  Same  Triangle  on  the  Meridian. — Let 
PZSN  be  the  meridian  (Fig.  74),  and  draw  the  diameters  ZWNa 
and  NWS  at  right  angles.  Lay  off  PZ&3Q°-L  and  draw  the 
diameters  PP'  and  QQ'  at  right  angles.  Since  the  object  is  west 
of  the  meridian,  the  center  of  the  primitive  circle  is  W. 


FIG.  74. 


From  Z,  lay  off  ZH  =  90°  —  h,  and  draw  a  tangent  at  H  inter- 
secting WZ  in  c,  the  center  of  the  small  circle  having  a  polar  dis- 
tance of  90° -h  (Art.  123,  Second  Method);  with  radius  cH 
draw  the  small  circle  HM. 

From  P  lay  off  Pr=90°  -d  =  58°,  and  draw  a  tangent  at  r, 
intersecting  PP^  in  cx,  the  center  of  the  small  circle  of  58°. 
With  radius  c^r  draw  the  small  circle  rM,  intersecting  HM  in  M, 
which  has  the  given  polar  distance  from  both  Z  and  P,  re- 
spectively. 


124 


STEREOGRAPHIC  PROJECTIONS. 


The  perpendicular  at  the  middle  point  of  the  straight  line  ZM 
intersects  NS,  the  line  of  centers  of  Z,  in  the  center  for  the  great 
circle  ZM.  Construct  the  great  circle  PM  in  a  similar  way,  giving 
PZM,  the  projection  of  the  given  triangle. 

The  value  of  the  angle  PZM  is  given  by  the  arc  of  the  primitive 
marked  Z  (about  64°),  which  measures  the  projected  arc  of  the 


horizon  Nm,  which  is  in  turn  the  measure  of  the  azimuth.  The 
hour  angle  t  is  measured  by  the  arc  of  the  primitive  marked  t,  the 
true  length  of  the  projected  arc  Q'm^  of  the  equator,  about  53°  50'. 
139.  To  Project  the  Same  Triangle  on  the  Equator. — In  the 
projection  (Fig.  75)  the  primitive  circle  is  the  celestial  equator  or 
equinoctial,  and  the  pole  P  is  its  center. "  Call  the  vertical  diam- 
eter QQ'  the  projection  of  the  meridian  and  lay  off  WA  =  9Q°  —  L 
^70°,  drawing  EA  cutting  QQ'  in  Z,  the  projected  zenith.  Find 


STEREOGRAPHIC  PROJECTIONS.  125 

cl9  the  center  of  the  small  circle  of  (90°  —  ft)  =50°  polar  dis- 
tance, and  construct  the  circle;  find  Pb,  the  projected  length  of 
the  polar  distance  PM  =  58°,  and  draw  the  small  circle  inter- 
secting the  first  one  in  M.  Find  the  line  of  centers  of  all  great 
circles  through  Z,  and  the  center  c  of  that  one  which  passes 
through  M.  The  construction  is  obvious.  The  value  of  t,  54°  30', 
is  measured  by  the  arc  of  the  primitive  circle  as  indicated.  The 
azimuth  Z  is  measured  by  the  true  length  of  the  intercepted  arc 
Nm  of  the  horizon  ENW,  indicated  by  Z  on  the  primitive  circle, 
about  64°. 

140.  Numerical  Solution  of  the  Triangle.  —  This  problem  is  the 
time  sight,  of  almost  daily  occurrence  in  the  navigation  of  a  ship 
at  sea.  The  altitude  of  the  observed  body  is  measured  with  the 
sextant,  the  latitude  is  assumed  as  known,  and  the  declination 
of  the  observed  body  is  taken  from  the  tables  of  the  Nautical 
Almanac. 

To  derive  the  formulae,  used  in  the  solution  of  the  triangle,  let 
t  =  A,  Z  =  B,  in  the  formulae  of  Arts.  88  and  89  (Sph.  Trig.), 


sin  o  sin  c 


sin  a  sin  c 
Then 


c=90°-L 


s  —  c  =  %(L  +  p  —  h)=sf  —  h 

where  s1 '  =  %(h  +  L  +  p)  ;  substituting  these  values  in  the  above 
formulae, 

sin2  \t  —  cos  s'  sin (s'  —  h)  sec  L  cosec  p. 
cos2  \Z  =  cos  s'  cos  (s'  —  p)secL  sec  h. 


126  STEREOGRAPHIC  PROJECTIONS. 

Note  that  p,  the  polar  distance,  is  used,  and  not  d. 

In  making  the  solution,  always  use  the  form  given  below. 

h=   40°  00'  sec  0.11575 

L=  20°  00'  sec  0.02701  sec     .02701 

p=   58°  00'  esc     .07158 

2s' =  118°  00' 


s'=   59°  00'  cos  9.71184  cos  9.71184 

s'-h=   19°  00'  sin  9.51264 

s'-p=     1°  00'  cos  9.99993 

2)9.32307 
*=   27°  18'  12"  sin  9.66153 


t=   54°  36'  24"  2(9.85453 

3h  38m  25.6s  \Z  32°  14'  34"  cos  9.92726 

Z  64°  29'     8" 

141.  Case  II :  Given  t,  d  and  L,  or  Two  Sides  and  the  Included 
Angle.— Take  L  =  39°  K,  £=2h20m=35°,  d=S°  K 

1.  On  the  Horizon :  Let  NESW  be  the  horizon  (Fig.  76),  and 
draw  NZS  and  WZE  at  right  angles. 

Find  ZP,  the  projected  length  of  90°  -L=EA  =  51°  ;  construct 
the  projection  of  the  equator,  Ws'E,  by  drawing  We  parallel  to 
ZA  (Art.  124),  intersecting  NS  at  c,  the  center,  giving  We,  the 
radius. 

The  pole  p  of  the  hour  circle  PM,  making  an  angle  of  35°  with 
PZ,  must  lie  on  the  equator  35°  from  the  pole  of  PZ  or  E;  hence 
lay  off  Ee  =  35°  and  draw  Pe,  cutting  the  equator  at  p.  Since  the 
poles  and  center  of  any  great  circle  are  in  the  same  straight  line 
as  the  center  of  the  primitive,  draw  Zpc^  meeting  the  line  of 
centers  at  clf  the  center  of  the  required  circle,  and  construct  the 
great  circle  PM.  From  its  pole  p  draw  pm,  making  the  true 
length  of  PM,  measured  by  p'm  =  90° -d  =  82°. 


STEREOGRAPHIC  PROJECTIONS, 


127 


Draw  ZMriLi  the  great  circle  through  Z  and  M,  completing  the 
construction  of  PZM,  and  find  its  pole  p19  laying  off  m1p1  =  90°. 

To  find  the  altitude  h,  draw  pji^  giving  mjil}  the  true  length 
of  the  projected  h  (m17lf)=45°  50'.  The  azimuth  Z  is  given 
directly  by  the  arc  NWm1  =  l2Q°. 

2.  The  same  triangle  projected  on  the  meridian  in  Fig.  77, 
where  PZ  —  90°  —  L  as  before.  Draw  the  equator  Q'WQ,  which  is 


FIG.  76. 


the  line  of  centers,  as  also  the  locus  of  the  poles,  of  all  great  circles 
through  P.  Find  the  pole  of  PM  making  an  angle  of  35°  with 
the  meridian  by  taking  P'a=35°,  giving  Wp  the  projected  length 
of  the  distance  between  the  poles  of  the  two  circles ;  find  its  center 
c  by  taking  P'&  =  70°,  twice  the  angle  P'Pc=35°.  From  the 
center  c,  draw  PtP'  and  lay  off  Pm  =  9Q° -d  =  82°.  Draw  pm 
cutting  the  great  circle  at  M;  PM— projected  length  of  Pm  =  82°. 


128 


STEREOGRAPH ic  PROJECTIONS. 


Draw  a  perpendicular  to  middle  point  of  the  chord  Z'fiQ inter- 
secting the  line  of  centers  of  great  circles  through  Z  at  c1?  the 
center  of  ZMm^Na.  Find  its  pole  pi  by  taking  ,szt  — 90°.  Draw 


FIG.  77. 

p^M,  giving  the  true  length  of  the  projected  altitude  Mm^  as 
indicated  approximately  45°  50'.  The  azimuth  PZM  is  measured 
by  the  true  length  of  the  projected  arc  of  the  horizon  N  Wm^ 
that  is,  by  the  arc  NQ'z,  about  125°. 

3.  The  same  triangle  is  projected  in  the  plane  of  the  equator  in 
Fig.  78,  where  SPN'  is  taken  as  the  projection  of  the  meridian, 
and  the  projection  of  the  zenith  found  by  making  PZ  =  9Q° 
-L.  Draw  PMt,  making  the  hour  angle  ZPM  =  35°,  and  find 
the  point  M  by  drawing  pM  from  pole  p  of  the  hour  circle  PM, 
cutting  off  Mt  =  S°  projective  length. 

Bisect  chord  ZM  as  before  by  a  perpendicular  meeting  the  line 
of  centers  at  c;  then  construct  ZM  meeting  horizon  produced  at 
m,  and  find  its  pole  p±  by  joining  cP. 

The  altitude  Mm  is  measured  by  A-^  according  to  usual 


STEREOGRAPHIC  PROJECTIONS. 


method,  and  is  about  45°  45'.    The  azimuth,  MZP,  is  measured  by 
the  true  length  of  the  arc  of  the  horizon  N Wm,  or  N'Wz 
nearly. 


142.  Numerical  Solution  of  Case  II. — By  Napier's  rules,  drop- 
ping the  perpendicular  from  M  on  PZ  : 
tan  <f>  =  cos  t  cot  d, 

sin/t=:sin^  sin(Z/  +  <£)sec  </>, 
cot  Z  =  cot  t  cos (L  +  <f>)  cosec  <£. 

Note  that  when  d  has  a  different  name  from  L,  d  is  negative. 

L   =  39°  00'  00" 

cos  9.91336  cot  0.15477 

cot  0.85220     sin  9.14356 


=  35°  00'  00" 
=     8°  00'  00" 


=  80°  15    53" 
=119°  15'  53" 

=  45°  53'  00" 

=125°  18'  40" 
10 


tan    0.76556 


sec  0.77178 
sin  9.94070 


esc     0.00630 
cos     9.68917n 


sin  9.85604 


cot    9.85024w 


130  STEREOGRAPHIO  PROJECTIONS. 

143.  Method  of  St.  Hilaire. — In  the  practice  of  navigation  the 
value  of  Z  is  taken  from  Azimuth  Tables,  and  h  is  derived  from 
the  formula  of  Art.  107  (Sph.  Trig.), 

sin  h  =  sin  d  sin  L+ cos  d  cos  L  cos  i, 
=cos(L—d)  —  2  cos  d  cos  L  sin2 1/. 

log  cos  L  9.89050          cos  (L-d)  0.85717  (Table  41) 
log  cos  d  9.99575  No.  0.13917 

log  2  0.30103 

0.18728  sinfc  0.71800  (Table  41) 

log  sin2  $t  8.95628  A- 45°  53'  23" 

No.  0.13917     log  9.14356 

The  computed  value  of  h  is  compared  with  the  observed  value, 
thus  furnishing  a  method  of  establishing  the  position  of  the  ship, 
instead  of  computing  t  by  Art.  140. 

Log  sin2^  may  be  taken  directly  from  Table  45  of  latest 
edition  of  Bowditch's  Useful  Tables. 

144.  Case  III :    Given  t,  d  and  h,  to  Find  L. — We  have  given 
two  sides  and  an  angle  (t)  opposite  co-h,  thus  furnishing  a  double 
solution.    Assume  t=32°,  d=21°  N.,  ^  =  40°  W.    The  projection 
is  made  on  the  equator,  on  account  of  the  simplicity  of  the  con- 
struction. 

Let  EQ'WQ  be  the  primitive  circle,  the  equator,  which  has  the 
pole  P  at  the  center.  Draw  the  diameter  QPQ',  and  EW  at  right 
angles  (Fig.  79).  Construct  the  angle  QPM =2=32°,  and  PM 
as  shown  in  the  figure  equal  to  90°  —  d=89°.  Prom  M  as  a  pole 
construct  the  small  circle  having  a  polar  distance  (90°  —  h)  =50°, 
on  which  lie  all  points  having  the  given  zenith  distance  of  50°. 
Then  the  two  points  Z^  and  Z2,  in  which  this  circle  of  equal  alti- 
tudes intersects  PQ,  satisfy  the  conditions  of  the  problem.  Find 
the  line  of  centers  for  M  and  construct  the  great  circles  Z^M  and 
Z2M,  completing  the  two  triangles. 


STEEEOGRAPHIC  PROJECTIONS. 


131 


Since  P  is  the  projection  of  the  North  Pole,  Z1?  lying  between 
it  and  the  equator,  is  the  projected  zenith  of  a  place  in  north  lati- 
tude, while  Z2,  lying  at  a  greater  polar  distance  than  90°,  is  the 
projected  zenith  of  a  place  in  south  latitude.  The  true  length  of 
Z^Q  is  measured  by  the  arc  marked  L1  =  66°  50'  N.;  the  true 
length  of  QZ2,  by  the  arc  marked  L2  =  18°  S. 


FIG.  79. 


145.  Numerical  Solution  of  Case  III. — Draw  the  perpendicular 
from  M  to  the  side  PZ±Z2)  and  by  Napier's  rules  derive : 

tan  (j>  =  cos  t  cot  d, 

cos  <£'  =  cos  <j>  sin  h  coseo  d, 

/,  =  90 -(<#>:±  </»'), 
cos  Z  =  tan  <f>'  tan  h. 


132 


STEREOGRAPHIC  PROJECTIONS. 


Note  that  <f>',  being  found  by  its  cosine,  may  be  either  a  positive 
a  negative  arc. 


or 

h  =  40°  00'  00" 

t  =  32°  00'  00" 

d  =  21°  00'  00"  N. 

0  =  65°  38'  46" 

0'=±  (42°  17'  55") 

Zl=  40°  13'  38" 

Zj=  139°  46'  22" 

L1=:  17°  56'  41"  S. 

L,=  66°  39'  9"  N. 


cos  9.92842 
cot  0.41582 


tan  0.34424 


sin  9.80807       tan 

esc  0.44567 
cos  9.61529 


9.92381 


cos  9.86903   tan  (±)  9.95899 


cos  (±)  9.88280 


AN  INITIAL  FINE  OF  25  CENTS 

OVERDUE. 


Berkeley 


yp    j  7 
ID    i 


'-.-.. 


3 G 0378 


UNIVERSITY  OF  CALIFORNIA  LIBRARY 


